Calculus Examples

$f (x) = - x^{4} + 2 x^{3} + 4 x^{2} - 2$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $- x^{4} + 2 x^{3} + 4 x^{2} - 2$ with respect to $x$ is $\frac{d}{d x} [- x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$ .

$\frac{d}{d x} [- x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Evaluate $\frac{d}{d x} [- x^{4}]$ .

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$- \frac{d}{d x} [x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$- (4 x^{3}) + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Multiply $4$ by $- 1$ .

$- 4 x^{3} + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$- 4 x^{3} + 2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$- 4 x^{3} + 2 (3 x^{2}) + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Multiply $3$ by $2$ .

$- 4 x^{3} + 6 x^{2} + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$- 4 x^{3} + 6 x^{2} + 4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$- 4 x^{3} + 6 x^{2} + 4 (2 x) + \frac{d}{d x} [- 2]$

Multiply $2$ by $4$ .

$- 4 x^{3} + 6 x^{2} + 8 x + \frac{d}{d x} [- 2]$

Differentiate using the Constant Rule.

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Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$- 4 x^{3} + 6 x^{2} + 8 x + 0$

Add $- 4 x^{3} + 6 x^{2} + 8 x$ and $0$ .

$- 4 x^{3} + 6 x^{2} + 8 x$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $- 4 x^{3} + 6 x^{2} + 8 x$ with respect to $x$ is $\frac{d}{d x} [- 4 x^{3}] + \frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [8 x]$ .

$f'' (x) = \frac{d}{d x} (- 4 x^{3}) + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (8 x)$

Evaluate $\frac{d}{d x} [- 4 x^{3}]$ .

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Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{3}$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = - 4 \frac{d}{d x} x^{3} + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (8 x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = - 4 (3 x^{2}) + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (8 x)$

Multiply $3$ by $- 4$ .

$f'' (x) = - 12 x^{2} + \frac{d}{d x} (6 x^{2}) + \frac{d}{d x} (8 x)$

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - 12 x^{2} + 6 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (8 x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = - 12 x^{2} + 6 (2 x) + \frac{d}{d x} (8 x)$

Multiply $2$ by $6$ .

$f'' (x) = - 12 x^{2} + 12 x + \frac{d}{d x} (8 x)$

Evaluate $\frac{d}{d x} [8 x]$ .

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Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f'' (x) = - 12 x^{2} + 12 x + 8 \frac{d}{d x} (x)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - 12 x^{2} + 12 x + 8 \cdot 1$

Multiply $8$ by $1$ .

$f'' (x) = - 12 x^{2} + 12 x + 8$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 4 x^{3} + 6 x^{2} + 8 x = 0$

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $- x^{4} + 2 x^{3} + 4 x^{2} - 2$ with respect to $x$ is $\frac{d}{d x} [- x^{4}] + \frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 2]$ .

$f' (x) = \frac{d}{d x} (- x^{4}) + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Evaluate $\frac{d}{d x} [- x^{4}]$ .

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Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4}$ with respect to $x$ is $- \frac{d}{d x} [x^{4}]$ .

$f' (x) = - \frac{d}{d x} x^{4} + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = - (4 x^{3}) + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Multiply $4$ by $- 1$ .

$f' (x) = - 4 x^{3} + \frac{d}{d x} (2 x^{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

$f' (x) = - 4 x^{3} + 2 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = - 4 x^{3} + 2 (3 x^{2}) + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Multiply $3$ by $2$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + \frac{d}{d x} (4 x^{2}) + \frac{d}{d x} (- 2)$

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + 4 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 2)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + 4 (2 x) + \frac{d}{d x} (- 2)$

Multiply $2$ by $4$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + 8 x + \frac{d}{d x} (- 2)$

Differentiate using the Constant Rule.

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Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + 8 x + 0$

Add $- 4 x^{3} + 6 x^{2} + 8 x$ and $0$ .

$f' (x) = - 4 x^{3} + 6 x^{2} + 8 x$

The first derivative of $f (x)$ with respect to $x$ is $- 4 x^{3} + 6 x^{2} + 8 x$ .

$- 4 x^{3} + 6 x^{2} + 8 x$

Set the first derivative equal to $0$ then solve the equation $- 4 x^{3} + 6 x^{2} + 8 x = 0$ .

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Set the first derivative equal to $0$ .

$- 4 x^{3} + 6 x^{2} + 8 x = 0$

Factor $- 2 x$ out of $- 4 x^{3} + 6 x^{2} + 8 x$ .

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Factor $- 2 x$ out of $- 4 x^{3}$ .

$- 2 x (2 x^{2}) + 6 x^{2} + 8 x = 0$

Factor $- 2 x$ out of $6 x^{2}$ .

$- 2 x (2 x^{2}) - 2 x (- 3 x) + 8 x = 0$

Factor $- 2 x$ out of $8 x$ .

$- 2 x (2 x^{2}) - 2 x (- 3 x) - 2 x \cdot - 4 = 0$

Factor $- 2 x$ out of $- 2 x (2 x^{2}) - 2 x (- 3 x)$ .

$- 2 x (2 x^{2} - 3 x) - 2 x \cdot - 4 = 0$

Factor $- 2 x$ out of $- 2 x (2 x^{2} - 3 x) - 2 x (- 4)$ .

$- 2 x (2 x^{2} - 3 x - 4) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x = 0$

$2 x^{2} - 3 x - 4 = 0$

Set $x$ equal to $0$ .

$x = 0$

Set $2 x^{2} - 3 x - 4$ equal to $0$ and solve for $x$ .

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Set $2 x^{2} - 3 x - 4$ equal to $0$ .

$2 x^{2} - 3 x - 4 = 0$

Solve $2 x^{2} - 3 x - 4 = 0$ for $x$ .

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 2$ , $b = - 3$ , and $c = - 4$ into the quadratic formula and solve for $x$ .

$\frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \cdot (2 \cdot - 4)}}{2 \cdot 2}$

Simplify.

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Simplify the numerator.

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Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 4}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 4}}{2 \cdot 2}$

Multiply $- 8$ by $- 4$ .

$x = \frac{3 \pm \sqrt{9 + 32}}{2 \cdot 2}$

Add $9$ and $32$ .

$x = \frac{3 \pm \sqrt{41}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{41}}{4}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 4}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 4}}{2 \cdot 2}$

Multiply $- 8$ by $- 4$ .

$x = \frac{3 \pm \sqrt{9 + 32}}{2 \cdot 2}$

Add $9$ and $32$ .

$x = \frac{3 \pm \sqrt{41}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{41}}{4}$

Change the $\pm$ to $+$ .

$x = \frac{3 + \sqrt{41}}{4}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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Raise $- 3$ to the power of $2$ .

$x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot - 4}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{3 \pm \sqrt{9 - 8 \cdot - 4}}{2 \cdot 2}$

Multiply $- 8$ by $- 4$ .

$x = \frac{3 \pm \sqrt{9 + 32}}{2 \cdot 2}$

Add $9$ and $32$ .

$x = \frac{3 \pm \sqrt{41}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{3 \pm \sqrt{41}}{4}$

Change the $\pm$ to $-$ .

$x = \frac{3 - \sqrt{41}}{4}$

The final answer is the combination of both solutions.

$x = \frac{3 + \sqrt{41}}{4}, \frac{3 - \sqrt{41}}{4}$

The final solution is all the values that make $- 2 x (2 x^{2} - 3 x - 4) = 0$ true.

$x = 0, \frac{3 + \sqrt{41}}{4}, \frac{3 - \sqrt{41}}{4}$

Find the values where the derivative is undefined.

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Critical points to evaluate.

$x = 0, \frac{3 + \sqrt{41}}{4}, \frac{3 - \sqrt{41}}{4}$

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(0)}^{2} + 12 (0) + 8$

Evaluate the second derivative.

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Simplify each term.

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Raising $0$ to any positive power yields $0$ .

$- 12 \cdot 0 + 12 (0) + 8$

Multiply $- 12$ by $0$ .

$0 + 12 (0) + 8$

Multiply $12$ by $0$ .

$0 + 0 + 8$

Simplify by adding numbers.

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Add $0$ and $0$ .

$0 + 8$

Add $0$ and $8$ .

$8$

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Find the y-value when $x = 0$ .

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Replace the variable $x$ with $0$ in the expression.

$f (0) = - {(0)}^{4} + 2 {(0)}^{3} + 4 {(0)}^{2} - 2$

Simplify the result.

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Simplify each term.

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Raising $0$ to any positive power yields $0$ .

$f (0) = - 0 + 2 {(0)}^{3} + 4 {(0)}^{2} - 2$

Multiply $- 1$ by $0$ .

$f (0) = 0 + 2 {(0)}^{3} + 4 {(0)}^{2} - 2$

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 2 \cdot 0 + 4 {(0)}^{2} - 2$

Multiply $2$ by $0$ .

$f (0) = 0 + 0 + 4 {(0)}^{2} - 2$

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 + 0 + 4 \cdot 0 - 2$

Multiply $4$ by $0$ .

$f (0) = 0 + 0 + 0 - 2$

Simplify by adding and subtracting.

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Add $0$ and $0$ .

$f (0) = 0 + 0 - 2$

Add $0$ and $0$ .

$f (0) = 0 - 2$

Subtract $2$ from $0$ .

$f (0) = - 2$

The final answer is $- 2$ .

$y = - 2$

Evaluate the second derivative at $x = \frac{3 + \sqrt{41}}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(\frac{3 + \sqrt{41}}{4})}^{2} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Evaluate the second derivative.

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Simplify each term.

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Apply the product rule to $\frac{3 + \sqrt{41}}{4}$ .

$- 12 \frac{{(3 + \sqrt{41})}^{2}}{4^{2}} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Raise $4$ to the power of $2$ .

$- 12 \frac{{(3 + \sqrt{41})}^{2}}{16} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factor of $4$ .

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Factor $4$ out of $- 12$ .

$4 (- 3) \frac{{(3 + \sqrt{41})}^{2}}{16} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Factor $4$ out of $16$ .

$4 \cdot - 3 \frac{{(3 + \sqrt{41})}^{2}}{4 \cdot 4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factor.

$4 \cdot - 3 \frac{{(3 + \sqrt{41})}^{2}}{4 \cdot 4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Rewrite the expression.

$- 3 \frac{{(3 + \sqrt{41})}^{2}}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Combine $- 3$ and $\frac{{(3 + \sqrt{41})}^{2}}{4}$ .

$\frac{- 3 {(3 + \sqrt{41})}^{2}}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Rewrite ${(3 + \sqrt{41})}^{2}$ as $(3 + \sqrt{41}) (3 + \sqrt{41})$ .

$\frac{- 3 ((3 + \sqrt{41}) (3 + \sqrt{41}))}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Expand $(3 + \sqrt{41}) (3 + \sqrt{41})$ using the FOIL Method.

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Apply the distributive property.

$\frac{- 3 (3 (3 + \sqrt{41}) + \sqrt{41} (3 + \sqrt{41}))}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Apply the distributive property.

$\frac{- 3 (3 \cdot 3 + 3 \sqrt{41} + \sqrt{41} (3 + \sqrt{41}))}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Apply the distributive property.

$\frac{- 3 (3 \cdot 3 + 3 \sqrt{41} + \sqrt{41} \cdot 3 + \sqrt{41} \sqrt{41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Simplify and combine like terms.

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Simplify each term.

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Multiply $3$ by $3$ .

$\frac{- 3 (9 + 3 \sqrt{41} + \sqrt{41} \cdot 3 + \sqrt{41} \sqrt{41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Move $3$ to the left of $\sqrt{41}$ .

$\frac{- 3 (9 + 3 \sqrt{41} + 3 \cdot \sqrt{41} + \sqrt{41} \sqrt{41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Combine using the product rule for radicals.

$\frac{- 3 (9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{41 \cdot 41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Multiply $41$ by $41$ .

$\frac{- 3 (9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{1681})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Rewrite $1681$ as $41^{2}$ .

$\frac{- 3 (9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{41^{2}})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{- 3 (9 + 3 \sqrt{41} + 3 \sqrt{41} + 41)}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Add $9$ and $41$ .

$\frac{- 3 (50 + 3 \sqrt{41} + 3 \sqrt{41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Add $3 \sqrt{41}$ and $3 \sqrt{41}$ .

$\frac{- 3 (50 + 6 \sqrt{41})}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factor of $50 + 6 \sqrt{41}$ and $4$ .

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Factor $2$ out of $- 3 (50 + 6 \sqrt{41})$ .

$\frac{2 (- 3 (25 + 3 \sqrt{41}))}{4} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factors.

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Factor $2$ out of $4$ .

$\frac{2 (- 3 (25 + 3 \sqrt{41}))}{2 (2)} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factor.

$\frac{2 (- 3 (25 + 3 \sqrt{41}))}{2 \cdot 2} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Rewrite the expression.

$\frac{- 3 (25 + 3 \sqrt{41})}{2} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Move the negative in front of the fraction.

$- \frac{(3) (25 + 3 \sqrt{41})}{2} + 12 (\frac{3 + \sqrt{41}}{4}) + 8$

Cancel the common factor of $4$ .

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Factor $4$ out of $12$ .

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 4 (3) \frac{3 + \sqrt{41}}{4} + 8$

Cancel the common factor.

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 4 \cdot 3 \frac{3 + \sqrt{41}}{4} + 8$

Rewrite the expression.

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 3 (3 + \sqrt{41}) + 8$

Apply the distributive property.

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 3 \cdot 3 + 3 \sqrt{41} + 8$

Multiply $3$ by $3$ .

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 9 + 3 \sqrt{41} + 8$

To write $9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{3 (25 + 3 \sqrt{41})}{2} + 9 \cdot \frac{2}{2} + 3 \sqrt{41} + 8$

Combine fractions.

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Combine $9$ and $\frac{2}{2}$ .

$- \frac{3 (25 + 3 \sqrt{41})}{2} + \frac{9 \cdot 2}{2} + 3 \sqrt{41} + 8$

Simplify the expression.

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Combine the numerators over the common denominator.

$\frac{- 3 (25 + 3 \sqrt{41}) + 9 \cdot 2}{2} + 3 \sqrt{41} + 8$

Multiply $9$ by $2$ .

$\frac{- 3 (25 + 3 \sqrt{41}) + 18}{2} + 3 \sqrt{41} + 8$

Simplify the numerator.

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Apply the distributive property.

$\frac{- 3 \cdot 25 - 3 (3 \sqrt{41}) + 18}{2} + 3 \sqrt{41} + 8$

Multiply $- 3$ by $25$ .

$\frac{- 75 - 3 (3 \sqrt{41}) + 18}{2} + 3 \sqrt{41} + 8$

Multiply $3$ by $- 3$ .

$\frac{- 75 - 9 \sqrt{41} + 18}{2} + 3 \sqrt{41} + 8$

Add $- 75$ and $18$ .

$\frac{- 57 - 9 \sqrt{41}}{2} + 3 \sqrt{41} + 8$

To write $3 \sqrt{41}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{- 57 - 9 \sqrt{41}}{2} + 3 \sqrt{41} \cdot \frac{2}{2} + 8$

Combine fractions.

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Combine $3 \sqrt{41}$ and $\frac{2}{2}$ .

$\frac{- 57 - 9 \sqrt{41}}{2} + \frac{3 \sqrt{41} \cdot 2}{2} + 8$

Combine the numerators over the common denominator.

$\frac{- 57 - 9 \sqrt{41} + 3 \sqrt{41} \cdot 2}{2} + 8$

Simplify the numerator.

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Multiply $2$ by $3$ .

$\frac{- 57 - 9 \sqrt{41} + 6 \sqrt{41}}{2} + 8$

Add $- 9 \sqrt{41}$ and $6 \sqrt{41}$ .

$\frac{- 57 - 3 \sqrt{41}}{2} + 8$

To write $8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{- 57 - 3 \sqrt{41}}{2} + 8 \cdot \frac{2}{2}$

Combine fractions.

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Combine $8$ and $\frac{2}{2}$ .

$\frac{- 57 - 3 \sqrt{41}}{2} + \frac{8 \cdot 2}{2}$

Combine the numerators over the common denominator.

$\frac{- 57 - 3 \sqrt{41} + 8 \cdot 2}{2}$

Simplify the numerator.

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Multiply $8$ by $2$ .

$\frac{- 57 - 3 \sqrt{41} + 16}{2}$

Add $- 57$ and $16$ .

$\frac{- 41 - 3 \sqrt{41}}{2}$

Simplify with factoring out.

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Rewrite $- 41$ as $- 1 (41)$ .

$\frac{- 1 (41) - 3 \sqrt{41}}{2}$

Factor $- 1$ out of $- 3 \sqrt{41}$ .

$\frac{- 1 (41) - (3 \sqrt{41})}{2}$

Factor $- 1$ out of $- 1 (41) - (3 \sqrt{41})$ .

$\frac{- 1 (41 + 3 \sqrt{41})}{2}$

Move the negative in front of the fraction.

$- \frac{41 + 3 \sqrt{41}}{2}$

$x = \frac{3 + \sqrt{41}}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{3 + \sqrt{41}}{4}$ is a local maximum

Find the y-value when $x = \frac{3 + \sqrt{41}}{4}$ .

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Replace the variable $x$ with $\frac{3 + \sqrt{41}}{4}$ in the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - {(\frac{3 + \sqrt{41}}{4})}^{4} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{3 + \sqrt{41}}{4}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{{(3 + \sqrt{41})}^{4}}{4^{4}} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $4$ to the power of $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{{(3 + \sqrt{41})}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Use the Binomial Theorem.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{3^{4} + 4 \cdot (3^{3} \sqrt{41}) + 6 \cdot (3^{2} {\sqrt{41}}^{2}) + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Simplify each term.

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Raise $3$ to the power of $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 4 \cdot (3^{3} \sqrt{41}) + 6 \cdot (3^{2} {\sqrt{41}}^{2}) + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 4 \cdot (27 \sqrt{41}) + 6 \cdot (3^{2} {\sqrt{41}}^{2}) + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $4$ by $27$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 6 \cdot (3^{2} {\sqrt{41}}^{2}) + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 6 \cdot (9 {\sqrt{41}}^{2}) + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $6$ by $9$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 54 {\sqrt{41}}^{2} + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 54 {(41^{\frac{1}{2}})}^{2} + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 54 \cdot 41^{\frac{1}{2} \cdot 2} + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 54 \cdot 41^{\frac{2}{2}} + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Cancel the common factor.

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 54 \cdot 41 + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Evaluate the exponent.

Multiply $54$ by $41$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 4 \cdot (3 {\sqrt{41}}^{3}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $4$ by $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 {\sqrt{41}}^{3} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{3}$ as ${(41^{3})}^{\frac{1}{2}}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 \sqrt{41^{3}} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 \sqrt{68921} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite $68921$ as $41^{2} \cdot 41$ .

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Factor $1681$ out of $68921$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 \sqrt{1681 (41)} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite $1681$ as $41^{2}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 \sqrt{41^{2} \cdot 41} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Pull terms out from under the radical.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 12 (41 \sqrt{41}) + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $41$ by $12$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{4}$ as $41^{2}$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + {(41^{\frac{1}{2}})}^{4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{1}{2} \cdot 4}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{4}{2}}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2}}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2 (1)}}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2 \cdot 1}}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{\frac{2}{1}}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Divide $2$ by $1$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 41^{2}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{81 + 108 \sqrt{41} + 2214 + 492 \sqrt{41} + 1681}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $81$ and $2214$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{2295 + 108 \sqrt{41} + 492 \sqrt{41} + 1681}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $2295$ and $1681$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{3976 + 108 \sqrt{41} + 492 \sqrt{41}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $108 \sqrt{41}$ and $492 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{3976 + 600 \sqrt{41}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $3976 + 600 \sqrt{41}$ and $256$ .

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Factor $8$ out of $3976$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{8 (497) + 600 \sqrt{41}}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Factor $8$ out of $600 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{8 (497) + 8 (75 \sqrt{41})}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Factor $8$ out of $8 (497) + 8 (75 \sqrt{41})$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{8 (497 + 75 \sqrt{41})}{256} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $8$ out of $256$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{8 (497 + 75 \sqrt{41})}{8 \cdot 32} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{8 (497 + 75 \sqrt{41})}{8 \cdot 32} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + 2 {(\frac{3 + \sqrt{41}}{4})}^{3} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $\frac{3 + \sqrt{41}}{4}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + 2 (\frac{{(3 + \sqrt{41})}^{3}}{4^{3}}) + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $4$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + 2 (\frac{{(3 + \sqrt{41})}^{3}}{64}) + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Factor $2$ out of $64$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + 2 (\frac{{(3 + \sqrt{41})}^{3}}{2 (32)}) + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + 2 (\frac{{(3 + \sqrt{41})}^{3}}{2 \cdot 32}) + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{{(3 + \sqrt{41})}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Use the Binomial Theorem.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{3^{3} + 3 \cdot (3^{2} \sqrt{41}) + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Simplify each term.

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Raise $3$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 3 \cdot (3^{2} \sqrt{41}) + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $3$ by $3^{2}$ by adding the exponents.

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Multiply $3$ by $3^{2}$ .

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Raise $3$ to the power of $1$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 3 \cdot (3^{2} \sqrt{41}) + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 3^{1 + 2} \sqrt{41} + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $1$ and $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 3^{3} \sqrt{41} + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 3 \cdot (3 {\sqrt{41}}^{2}) + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $3$ by $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 {\sqrt{41}}^{2} + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 {(41^{\frac{1}{2}})}^{2} + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 \cdot 41^{\frac{1}{2} \cdot 2} + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 \cdot 41^{\frac{2}{2}} + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 \cdot 41^{\frac{2}{2}} + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 \cdot 41 + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Evaluate the exponent.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 9 \cdot 41 + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Multiply $9$ by $41$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{3}$ as ${(41^{3})}^{\frac{1}{2}}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + \sqrt{41^{3}}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + \sqrt{68921}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite $68921$ as $41^{2} \cdot 41$ .

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Factor $1681$ out of $68921$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + \sqrt{1681 (41)}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite $1681$ as $41^{2}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + \sqrt{41^{2} \cdot 41}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Pull terms out from under the radical.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{27 + 27 \sqrt{41} + 369 + 41 \sqrt{41}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $27$ and $369$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{396 + 27 \sqrt{41} + 41 \sqrt{41}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Add $27 \sqrt{41}$ and $41 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{396 + 68 \sqrt{41}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $396 + 68 \sqrt{41}$ and $32$ .

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Factor $4$ out of $396$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{4 (99) + 68 \sqrt{41}}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Factor $4$ out of $68 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{4 (99) + 4 (17 \sqrt{41})}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Factor $4$ out of $4 (99) + 4 (17 \sqrt{41})$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{4 (99 + 17 \sqrt{41})}{32} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $4$ out of $32$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{4 (99 + 17 \sqrt{41})}{4 \cdot 8} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{4 (99 + 17 \sqrt{41})}{4 \cdot 8} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + 4 {(\frac{3 + \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $\frac{3 + \sqrt{41}}{4}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + 4 (\frac{{(3 + \sqrt{41})}^{2}}{4^{2}}) - 2$

Raise $4$ to the power of $2$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + 4 (\frac{{(3 + \sqrt{41})}^{2}}{16}) - 2$

Cancel the common factor of $4$ .

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Factor $4$ out of $16$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + 4 (\frac{{(3 + \sqrt{41})}^{2}}{4 (4)}) - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + 4 (\frac{{(3 + \sqrt{41})}^{2}}{4 \cdot 4}) - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{{(3 + \sqrt{41})}^{2}}{4} - 2$

Rewrite ${(3 + \sqrt{41})}^{2}$ as $(3 + \sqrt{41}) (3 + \sqrt{41})$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{(3 + \sqrt{41}) (3 + \sqrt{41})}{4} - 2$

Expand $(3 + \sqrt{41}) (3 + \sqrt{41})$ using the FOIL Method.

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Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{3 (3 + \sqrt{41}) + \sqrt{41} (3 + \sqrt{41})}{4} - 2$

Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{3 \cdot 3 + 3 \sqrt{41} + \sqrt{41} (3 + \sqrt{41})}{4} - 2$

Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{3 \cdot 3 + 3 \sqrt{41} + \sqrt{41} \cdot 3 + \sqrt{41} \sqrt{41}}{4} - 2$

Simplify and combine like terms.

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Simplify each term.

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Multiply $3$ by $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + \sqrt{41} \cdot 3 + \sqrt{41} \sqrt{41}}{4} - 2$

Move $3$ to the left of $\sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + 3 \cdot \sqrt{41} + \sqrt{41} \sqrt{41}}{4} - 2$

Combine using the product rule for radicals.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{41 \cdot 41}}{4} - 2$

Multiply $41$ by $41$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{1681}}{4} - 2$

Rewrite $1681$ as $41^{2}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + 3 \sqrt{41} + \sqrt{41^{2}}}{4} - 2$

Pull terms out from under the radical, assuming positive real numbers.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{9 + 3 \sqrt{41} + 3 \sqrt{41} + 41}{4} - 2$

Add $9$ and $41$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{50 + 3 \sqrt{41} + 3 \sqrt{41}}{4} - 2$

Add $3 \sqrt{41}$ and $3 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{50 + 6 \sqrt{41}}{4} - 2$

Cancel the common factor of $50 + 6 \sqrt{41}$ and $4$ .

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Factor $2$ out of $50$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{2 (25) + 6 \sqrt{41}}{4} - 2$

Factor $2$ out of $6 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{2 (25) + 2 (3 \sqrt{41})}{4} - 2$

Factor $2$ out of $2 (25) + 2 (3 \sqrt{41})$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{2 (25 + 3 \sqrt{41})}{4} - 2$

Cancel the common factors.

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Factor $2$ out of $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{2 (25 + 3 \sqrt{41})}{2 \cdot 2} - 2$

Cancel the common factor.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{2 (25 + 3 \sqrt{41})}{2 \cdot 2} - 2$

Rewrite the expression.

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} + \frac{25 + 3 \sqrt{41}}{2} - 2$

Find the common denominator.

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Multiply $\frac{99 + 17 \sqrt{41}}{8}$ by $\frac{4}{4}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{99 + 17 \sqrt{41}}{8} \cdot \frac{4}{4} + \frac{25 + 3 \sqrt{41}}{2} - 2$

Multiply $\frac{99 + 17 \sqrt{41}}{8}$ by $\frac{4}{4}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{25 + 3 \sqrt{41}}{2} - 2$

Multiply $\frac{25 + 3 \sqrt{41}}{2}$ by $\frac{16}{16}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{25 + 3 \sqrt{41}}{2} \cdot \frac{16}{16} - 2$

Multiply $\frac{25 + 3 \sqrt{41}}{2}$ by $\frac{16}{16}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} - 2$

Write $- 2$ as a fraction with denominator $1$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2}{1}$

Multiply $\frac{- 2}{1}$ by $\frac{32}{32}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2}{1} \cdot \frac{32}{32}$

Multiply $\frac{- 2}{1}$ by $\frac{32}{32}$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Reorder the factors of $8 \cdot 4$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{4 \cdot 8} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Multiply $4$ by $8$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{32} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Multiply $2$ by $16$ .

$f (\frac{3 + \sqrt{41}}{4}) = - \frac{497 + 75 \sqrt{41}}{32} + \frac{(99 + 17 \sqrt{41}) \cdot 4}{32} + \frac{(25 + 3 \sqrt{41}) \cdot 16}{32} + \frac{- 2 \cdot 32}{32}$

Combine the numerators over the common denominator.

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- (497 + 75 \sqrt{41}) + (99 + 17 \sqrt{41}) \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Simplify each term.

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Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 1 \cdot 497 - (75 \sqrt{41}) + (99 + 17 \sqrt{41}) \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $- 1$ by $497$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - (75 \sqrt{41}) + (99 + 17 \sqrt{41}) \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $75$ by $- 1$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + (99 + 17 \sqrt{41}) \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 99 \cdot 4 + 17 \sqrt{41} \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $99$ by $4$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 17 \sqrt{41} \cdot 4 + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $4$ by $17$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 68 \sqrt{41} + (25 + 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Apply the distributive property.

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 68 \sqrt{41} + 25 \cdot 16 + 3 \sqrt{41} \cdot 16 - 2 \cdot 32}{32}$

Multiply $25$ by $16$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 68 \sqrt{41} + 400 + 3 \sqrt{41} \cdot 16 - 2 \cdot 32}{32}$

Multiply $16$ by $3$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 68 \sqrt{41} + 400 + 48 \sqrt{41} - 2 \cdot 32}{32}$

Multiply $- 2$ by $32$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 497 - 75 \sqrt{41} + 396 + 68 \sqrt{41} + 400 + 48 \sqrt{41} - 64}{32}$

Simplify by adding terms.

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Add $- 497$ and $396$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{- 101 - 75 \sqrt{41} + 68 \sqrt{41} + 400 + 48 \sqrt{41} - 64}{32}$

Simplify by adding and subtracting.

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Add $- 101$ and $400$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{299 - 75 \sqrt{41} + 68 \sqrt{41} + 48 \sqrt{41} - 64}{32}$

Subtract $64$ from $299$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{235 - 75 \sqrt{41} + 68 \sqrt{41} + 48 \sqrt{41}}{32}$

Add $- 75 \sqrt{41}$ and $68 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{235 - 7 \sqrt{41} + 48 \sqrt{41}}{32}$

Add $- 7 \sqrt{41}$ and $48 \sqrt{41}$ .

$f (\frac{3 + \sqrt{41}}{4}) = \frac{235 + 41 \sqrt{41}}{32}$

The final answer is $\frac{235 + 41 \sqrt{41}}{32}$ .

$y = \frac{235 + 41 \sqrt{41}}{32}$

Evaluate the second derivative at $x = \frac{3 - \sqrt{41}}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 12 {(\frac{3 - \sqrt{41}}{4})}^{2} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Evaluate the second derivative.

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Simplify each term.

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Apply the product rule to $\frac{3 - \sqrt{41}}{4}$ .

$- 12 \frac{{(3 - \sqrt{41})}^{2}}{4^{2}} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Raise $4$ to the power of $2$ .

$- 12 \frac{{(3 - \sqrt{41})}^{2}}{16} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor of $4$ .

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Factor $4$ out of $- 12$ .

$4 (- 3) \frac{{(3 - \sqrt{41})}^{2}}{16} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Factor $4$ out of $16$ .

$4 \cdot - 3 \frac{{(3 - \sqrt{41})}^{2}}{4 \cdot 4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor.

$4 \cdot - 3 \frac{{(3 - \sqrt{41})}^{2}}{4 \cdot 4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Rewrite the expression.

$- 3 \frac{{(3 - \sqrt{41})}^{2}}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Combine $- 3$ and $\frac{{(3 - \sqrt{41})}^{2}}{4}$ .

$\frac{- 3 {(3 - \sqrt{41})}^{2}}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Rewrite ${(3 - \sqrt{41})}^{2}$ as $(3 - \sqrt{41}) (3 - \sqrt{41})$ .

$\frac{- 3 ((3 - \sqrt{41}) (3 - \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Expand $(3 - \sqrt{41}) (3 - \sqrt{41})$ using the FOIL Method.

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Apply the distributive property.

$\frac{- 3 (3 (3 - \sqrt{41}) - \sqrt{41} (3 - \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Apply the distributive property.

$\frac{- 3 (3 \cdot 3 + 3 (- \sqrt{41}) - \sqrt{41} (3 - \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Apply the distributive property.

$\frac{- 3 (3 \cdot 3 + 3 (- \sqrt{41}) - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Simplify and combine like terms.

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Simplify each term.

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Multiply $3$ by $3$ .

$\frac{- 3 (9 + 3 (- \sqrt{41}) - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Multiply $- 1$ by $3$ .

$\frac{- 3 (9 - 3 \sqrt{41} - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Multiply $3$ by $- 1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} - \sqrt{41} (- \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Multiply $- \sqrt{41} (- \sqrt{41})$ .

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Multiply $- 1$ by $- 1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 1 \sqrt{41} \sqrt{41})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Multiply $\sqrt{41}$ by $1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + \sqrt{41} \sqrt{41})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Raise $\sqrt{41}$ to the power of $1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{1} \sqrt{41})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Raise $\sqrt{41}$ to the power of $1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{1} {\sqrt{41}}^{1})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{1 + 1})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Add $1$ and $1$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{2})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + {(41^{\frac{1}{2}})}^{2})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{1}{2} \cdot 2})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Combine $\frac{1}{2}$ and $2$ .

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{2}{2}})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{2}{2}})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Rewrite the expression.

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{1})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Evaluate the exponent.

$\frac{- 3 (9 - 3 \sqrt{41} - 3 \sqrt{41} + 41)}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Add $9$ and $41$ .

$\frac{- 3 (50 - 3 \sqrt{41} - 3 \sqrt{41})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Subtract $3 \sqrt{41}$ from $- 3 \sqrt{41}$ .

$\frac{- 3 (50 - 6 \sqrt{41})}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor of $50 - 6 \sqrt{41}$ and $4$ .

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Factor $2$ out of $- 3 (50 - 6 \sqrt{41})$ .

$\frac{2 (- 3 (25 - 3 \sqrt{41}))}{4} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factors.

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Factor $2$ out of $4$ .

$\frac{2 (- 3 (25 - 3 \sqrt{41}))}{2 (2)} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor.

$\frac{2 (- 3 (25 - 3 \sqrt{41}))}{2 \cdot 2} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Rewrite the expression.

$\frac{- 3 (25 - 3 \sqrt{41})}{2} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Move the negative in front of the fraction.

$- \frac{(3) (25 - 3 \sqrt{41})}{2} + 12 (\frac{3 - \sqrt{41}}{4}) + 8$

Cancel the common factor of $4$ .

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Factor $4$ out of $12$ .

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 4 (3) \frac{3 - \sqrt{41}}{4} + 8$

Cancel the common factor.

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 4 \cdot 3 \frac{3 - \sqrt{41}}{4} + 8$

Rewrite the expression.

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 3 (3 - \sqrt{41}) + 8$

Apply the distributive property.

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 3 \cdot 3 + 3 (- \sqrt{41}) + 8$

Multiply $3$ by $3$ .

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 9 + 3 (- \sqrt{41}) + 8$

Multiply $- 1$ by $3$ .

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 9 - 3 \sqrt{41} + 8$

To write $9$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$- \frac{3 (25 - 3 \sqrt{41})}{2} + 9 \cdot \frac{2}{2} - 3 \sqrt{41} + 8$

Combine fractions.

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Combine $9$ and $\frac{2}{2}$ .

$- \frac{3 (25 - 3 \sqrt{41})}{2} + \frac{9 \cdot 2}{2} - 3 \sqrt{41} + 8$

Simplify the expression.

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Combine the numerators over the common denominator.

$\frac{- 3 (25 - 3 \sqrt{41}) + 9 \cdot 2}{2} - 3 \sqrt{41} + 8$

Multiply $9$ by $2$ .

$\frac{- 3 (25 - 3 \sqrt{41}) + 18}{2} - 3 \sqrt{41} + 8$

Simplify the numerator.

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Apply the distributive property.

$\frac{- 3 \cdot 25 - 3 (- 3 \sqrt{41}) + 18}{2} - 3 \sqrt{41} + 8$

Multiply $- 3$ by $25$ .

$\frac{- 75 - 3 (- 3 \sqrt{41}) + 18}{2} - 3 \sqrt{41} + 8$

Multiply $- 3$ by $- 3$ .

$\frac{- 75 + 9 \sqrt{41} + 18}{2} - 3 \sqrt{41} + 8$

Add $- 75$ and $18$ .

$\frac{- 57 + 9 \sqrt{41}}{2} - 3 \sqrt{41} + 8$

To write $- 3 \sqrt{41}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{- 57 + 9 \sqrt{41}}{2} - 3 \sqrt{41} \cdot \frac{2}{2} + 8$

Combine fractions.

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Combine $- 3 \sqrt{41}$ and $\frac{2}{2}$ .

$\frac{- 57 + 9 \sqrt{41}}{2} + \frac{- 3 \sqrt{41} \cdot 2}{2} + 8$

Combine the numerators over the common denominator.

$\frac{- 57 + 9 \sqrt{41} - 3 \sqrt{41} \cdot 2}{2} + 8$

Simplify the numerator.

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Multiply $2$ by $- 3$ .

$\frac{- 57 + 9 \sqrt{41} - 6 \sqrt{41}}{2} + 8$

Subtract $6 \sqrt{41}$ from $9 \sqrt{41}$ .

$\frac{- 57 + 3 \sqrt{41}}{2} + 8$

To write $8$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{- 57 + 3 \sqrt{41}}{2} + 8 \cdot \frac{2}{2}$

Combine fractions.

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Combine $8$ and $\frac{2}{2}$ .

$\frac{- 57 + 3 \sqrt{41}}{2} + \frac{8 \cdot 2}{2}$

Combine the numerators over the common denominator.

$\frac{- 57 + 3 \sqrt{41} + 8 \cdot 2}{2}$

Simplify the numerator.

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Multiply $8$ by $2$ .

$\frac{- 57 + 3 \sqrt{41} + 16}{2}$

Add $- 57$ and $16$ .

$\frac{- 41 + 3 \sqrt{41}}{2}$

Simplify with factoring out.

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Rewrite $- 41$ as $- 1 (41)$ .

$\frac{- 1 (41) + 3 \sqrt{41}}{2}$

Factor $- 1$ out of $3 \sqrt{41}$ .

$\frac{- 1 (41) - (- 3 \sqrt{41})}{2}$

Factor $- 1$ out of $- 1 (41) - (- 3 \sqrt{41})$ .

$\frac{- 1 (41 - 3 \sqrt{41})}{2}$

Move the negative in front of the fraction.

$- \frac{41 - 3 \sqrt{41}}{2}$

$x = \frac{3 - \sqrt{41}}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{3 - \sqrt{41}}{4}$ is a local maximum

Find the y-value when $x = \frac{3 - \sqrt{41}}{4}$ .

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Replace the variable $x$ with $\frac{3 - \sqrt{41}}{4}$ in the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - {(\frac{3 - \sqrt{41}}{4})}^{4} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Simplify the result.

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Simplify each term.

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Apply the product rule to $\frac{3 - \sqrt{41}}{4}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{{(3 - \sqrt{41})}^{4}}{4^{4}} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $4$ to the power of $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{{(3 - \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Use the Binomial Theorem.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{3^{4} + 4 \cdot (3^{3} (- \sqrt{41})) + 6 \cdot (3^{2} {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Simplify each term.

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Raise $3$ to the power of $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 + 4 \cdot (3^{3} (- \sqrt{41})) + 6 \cdot (3^{2} {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 + 4 \cdot (27 (- \sqrt{41})) + 6 \cdot (3^{2} {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $4$ by $27$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 + 108 (- \sqrt{41}) + 6 \cdot (3^{2} {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $- 1$ by $108$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 6 \cdot (3^{2} {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 6 \cdot (9 {(- \sqrt{41})}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $6$ by $9$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 {(- \sqrt{41})}^{2} + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $- \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 ({(- 1)}^{2} {\sqrt{41}}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $- 1$ to the power of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 (1 {\sqrt{41}}^{2}) + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply ${\sqrt{41}}^{2}$ by $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 {\sqrt{41}}^{2} + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 {(41^{\frac{1}{2}})}^{2} + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 \cdot 41^{\frac{1}{2} \cdot 2} + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 \cdot 41^{\frac{2}{2}} + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Cancel the common factor.

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 54 \cdot 41 + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Evaluate the exponent.

Multiply $54$ by $41$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 4 \cdot (3 {(- \sqrt{41})}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $4$ by $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 {(- \sqrt{41})}^{3} + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $- \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 ({(- 1)}^{3} {\sqrt{41}}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $- 1$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- {\sqrt{41}}^{3}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{3}$ as ${(41^{3})}^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- \sqrt{41^{3}}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- \sqrt{68921}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite $68921$ as $41^{2} \cdot 41$ .

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Factor $1681$ out of $68921$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- \sqrt{1681 (41)}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite $1681$ as $41^{2}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- \sqrt{41^{2} \cdot 41}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Pull terms out from under the radical.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- (41 \sqrt{41})) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $41$ by $- 1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 + 12 (- 41 \sqrt{41}) + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $- 41$ by $12$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + {(- \sqrt{41})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $- \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + {(- 1)}^{4} {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $- 1$ to the power of $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 1 {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply ${\sqrt{41}}^{4}$ by $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + {\sqrt{41}}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{4}$ as $41^{2}$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + {(41^{\frac{1}{2}})}^{4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{1}{2} \cdot 4}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{4}{2}}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2}}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2 (1)}}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{2 \cdot 2}{2 \cdot 1}}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{\frac{2}{1}}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Divide $2$ by $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 41^{2}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{81 - 108 \sqrt{41} + 2214 - 492 \sqrt{41} + 1681}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Add $81$ and $2214$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{2295 - 108 \sqrt{41} - 492 \sqrt{41} + 1681}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Add $2295$ and $1681$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{3976 - 108 \sqrt{41} - 492 \sqrt{41}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Subtract $492 \sqrt{41}$ from $- 108 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{3976 - 600 \sqrt{41}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $3976 - 600 \sqrt{41}$ and $256$ .

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Factor $8$ out of $3976$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{8 (497) - 600 \sqrt{41}}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Factor $8$ out of $- 600 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{8 (497) + 8 (- 75 \sqrt{41})}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Factor $8$ out of $8 (497) + 8 (- 75 \sqrt{41})$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{8 (497 - 75 \sqrt{41})}{256} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $8$ out of $256$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{8 (497 - 75 \sqrt{41})}{8 \cdot 32} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{8 (497 - 75 \sqrt{41})}{8 \cdot 32} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + 2 {(\frac{3 - \sqrt{41}}{4})}^{3} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $\frac{3 - \sqrt{41}}{4}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + 2 (\frac{{(3 - \sqrt{41})}^{3}}{4^{3}}) + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $4$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + 2 (\frac{{(3 - \sqrt{41})}^{3}}{64}) + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Factor $2$ out of $64$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + 2 (\frac{{(3 - \sqrt{41})}^{3}}{2 (32)}) + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + 2 (\frac{{(3 - \sqrt{41})}^{3}}{2 \cdot 32}) + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{{(3 - \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Use the Binomial Theorem.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{3^{3} + 3 \cdot (3^{2} (- \sqrt{41})) + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Simplify each term.

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Raise $3$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 + 3 \cdot (3^{2} (- \sqrt{41})) + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $3$ by $3^{2}$ by adding the exponents.

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Multiply $3$ by $3^{2}$ .

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Raise $3$ to the power of $1$ .

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 + 3^{1 + 2} (- \sqrt{41}) + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Add $1$ and $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 + 3^{3} (- \sqrt{41}) + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $3$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 + 27 (- \sqrt{41}) + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $- 1$ by $27$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 3 \cdot (3 {(- \sqrt{41})}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $3$ by $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 {(- \sqrt{41})}^{2} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $- \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 ({(- 1)}^{2} {\sqrt{41}}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $- 1$ to the power of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 (1 {\sqrt{41}}^{2}) + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply ${\sqrt{41}}^{2}$ by $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 {\sqrt{41}}^{2} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 {(41^{\frac{1}{2}})}^{2} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 \cdot 41^{\frac{1}{2} \cdot 2} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 \cdot 41^{\frac{2}{2}} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 \cdot 41^{\frac{2}{2}} + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 \cdot 41 + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Evaluate the exponent.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 9 \cdot 41 + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $9$ by $41$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 + {(- \sqrt{41})}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $- \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 + {(- 1)}^{3} {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $- 1$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - {\sqrt{41}}^{3}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite ${\sqrt{41}}^{3}$ as ${(41^{3})}^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - \sqrt{41^{3}}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Raise $41$ to the power of $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - \sqrt{68921}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite $68921$ as $41^{2} \cdot 41$ .

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Factor $1681$ out of $68921$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - \sqrt{1681 (41)}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite $1681$ as $41^{2}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - \sqrt{41^{2} \cdot 41}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Pull terms out from under the radical.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - (41 \sqrt{41})}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Multiply $41$ by $- 1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{27 - 27 \sqrt{41} + 369 - 41 \sqrt{41}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Add $27$ and $369$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{396 - 27 \sqrt{41} - 41 \sqrt{41}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Subtract $41 \sqrt{41}$ from $- 27 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{396 - 68 \sqrt{41}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor of $396 - 68 \sqrt{41}$ and $32$ .

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Factor $4$ out of $396$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{4 (99) - 68 \sqrt{41}}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Factor $4$ out of $- 68 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{4 (99) + 4 (- 17 \sqrt{41})}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Factor $4$ out of $4 (99) + 4 (- 17 \sqrt{41})$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{4 (99 - 17 \sqrt{41})}{32} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factors.

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Factor $4$ out of $32$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{4 (99 - 17 \sqrt{41})}{4 \cdot 8} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{4 (99 - 17 \sqrt{41})}{4 \cdot 8} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + 4 {(\frac{3 - \sqrt{41}}{4})}^{2} - 2$

Apply the product rule to $\frac{3 - \sqrt{41}}{4}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + 4 (\frac{{(3 - \sqrt{41})}^{2}}{4^{2}}) - 2$

Raise $4$ to the power of $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + 4 (\frac{{(3 - \sqrt{41})}^{2}}{16}) - 2$

Cancel the common factor of $4$ .

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Factor $4$ out of $16$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + 4 (\frac{{(3 - \sqrt{41})}^{2}}{4 (4)}) - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + 4 (\frac{{(3 - \sqrt{41})}^{2}}{4 \cdot 4}) - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{{(3 - \sqrt{41})}^{2}}{4} - 2$

Rewrite ${(3 - \sqrt{41})}^{2}$ as $(3 - \sqrt{41}) (3 - \sqrt{41})$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{(3 - \sqrt{41}) (3 - \sqrt{41})}{4} - 2$

Expand $(3 - \sqrt{41}) (3 - \sqrt{41})$ using the FOIL Method.

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Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{3 (3 - \sqrt{41}) - \sqrt{41} (3 - \sqrt{41})}{4} - 2$

Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{3 \cdot 3 + 3 (- \sqrt{41}) - \sqrt{41} (3 - \sqrt{41})}{4} - 2$

Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{3 \cdot 3 + 3 (- \sqrt{41}) - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41})}{4} - 2$

Simplify and combine like terms.

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Simplify each term.

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Multiply $3$ by $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 + 3 (- \sqrt{41}) - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41})}{4} - 2$

Multiply $- 1$ by $3$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - \sqrt{41} \cdot 3 - \sqrt{41} (- \sqrt{41})}{4} - 2$

Multiply $3$ by $- 1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} - \sqrt{41} (- \sqrt{41})}{4} - 2$

Multiply $- \sqrt{41} (- \sqrt{41})$ .

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Multiply $- 1$ by $- 1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 1 \sqrt{41} \sqrt{41}}{4} - 2$

Multiply $\sqrt{41}$ by $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + \sqrt{41} \sqrt{41}}{4} - 2$

Raise $\sqrt{41}$ to the power of $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + \sqrt{41} \sqrt{41}}{4} - 2$

Raise $\sqrt{41}$ to the power of $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + \sqrt{41} \sqrt{41}}{4} - 2$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{1 + 1}}{4} - 2$

Add $1$ and $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + {\sqrt{41}}^{2}}{4} - 2$

Rewrite ${\sqrt{41}}^{2}$ as $41$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{41}$ as $41^{\frac{1}{2}}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + {(41^{\frac{1}{2}})}^{2}}{4} - 2$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{1}{2} \cdot 2}}{4} - 2$

Combine $\frac{1}{2}$ and $2$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{2}{2}}}{4} - 2$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 41^{\frac{2}{2}}}{4} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 41}{4} - 2$

Evaluate the exponent.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{9 - 3 \sqrt{41} - 3 \sqrt{41} + 41}{4} - 2$

Add $9$ and $41$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{50 - 3 \sqrt{41} - 3 \sqrt{41}}{4} - 2$

Subtract $3 \sqrt{41}$ from $- 3 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{50 - 6 \sqrt{41}}{4} - 2$

Cancel the common factor of $50 - 6 \sqrt{41}$ and $4$ .

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Factor $2$ out of $50$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{2 (25) - 6 \sqrt{41}}{4} - 2$

Factor $2$ out of $- 6 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{2 (25) + 2 (- 3 \sqrt{41})}{4} - 2$

Factor $2$ out of $2 (25) + 2 (- 3 \sqrt{41})$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{2 (25 - 3 \sqrt{41})}{4} - 2$

Cancel the common factors.

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Factor $2$ out of $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{2 (25 - 3 \sqrt{41})}{2 \cdot 2} - 2$

Cancel the common factor.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{2 (25 - 3 \sqrt{41})}{2 \cdot 2} - 2$

Rewrite the expression.

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} + \frac{25 - 3 \sqrt{41}}{2} - 2$

Find the common denominator.

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Multiply $\frac{99 - 17 \sqrt{41}}{8}$ by $\frac{4}{4}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{99 - 17 \sqrt{41}}{8} \cdot \frac{4}{4} + \frac{25 - 3 \sqrt{41}}{2} - 2$

Multiply $\frac{99 - 17 \sqrt{41}}{8}$ by $\frac{4}{4}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{25 - 3 \sqrt{41}}{2} - 2$

Multiply $\frac{25 - 3 \sqrt{41}}{2}$ by $\frac{16}{16}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{25 - 3 \sqrt{41}}{2} \cdot \frac{16}{16} - 2$

Multiply $\frac{25 - 3 \sqrt{41}}{2}$ by $\frac{16}{16}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} - 2$

Write $- 2$ as a fraction with denominator $1$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2}{1}$

Multiply $\frac{- 2}{1}$ by $\frac{32}{32}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2}{1} \cdot \frac{32}{32}$

Multiply $\frac{- 2}{1}$ by $\frac{32}{32}$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{8 \cdot 4} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Reorder the factors of $8 \cdot 4$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{4 \cdot 8} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Multiply $4$ by $8$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{32} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{2 \cdot 16} + \frac{- 2 \cdot 32}{32}$

Multiply $2$ by $16$ .

$f (\frac{3 - \sqrt{41}}{4}) = - \frac{497 - 75 \sqrt{41}}{32} + \frac{(99 - 17 \sqrt{41}) \cdot 4}{32} + \frac{(25 - 3 \sqrt{41}) \cdot 16}{32} + \frac{- 2 \cdot 32}{32}$

Combine the numerators over the common denominator.

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- (497 - 75 \sqrt{41}) + (99 - 17 \sqrt{41}) \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Simplify each term.

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Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 1 \cdot 497 - (- 75 \sqrt{41}) + (99 - 17 \sqrt{41}) \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $- 1$ by $497$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 - (- 75 \sqrt{41}) + (99 - 17 \sqrt{41}) \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $- 75$ by $- 1$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + (99 - 17 \sqrt{41}) \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 99 \cdot 4 - 17 \sqrt{41} \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $99$ by $4$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 17 \sqrt{41} \cdot 4 + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Multiply $4$ by $- 17$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 68 \sqrt{41} + (25 - 3 \sqrt{41}) \cdot 16 - 2 \cdot 32}{32}$

Apply the distributive property.

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 68 \sqrt{41} + 25 \cdot 16 - 3 \sqrt{41} \cdot 16 - 2 \cdot 32}{32}$

Multiply $25$ by $16$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 68 \sqrt{41} + 400 - 3 \sqrt{41} \cdot 16 - 2 \cdot 32}{32}$

Multiply $16$ by $- 3$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 68 \sqrt{41} + 400 - 48 \sqrt{41} - 2 \cdot 32}{32}$

Multiply $- 2$ by $32$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 497 + 75 \sqrt{41} + 396 - 68 \sqrt{41} + 400 - 48 \sqrt{41} - 64}{32}$

Simplify by adding terms.

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Add $- 497$ and $396$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{- 101 + 75 \sqrt{41} - 68 \sqrt{41} + 400 - 48 \sqrt{41} - 64}{32}$

Simplify by adding and subtracting.

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Add $- 101$ and $400$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{299 + 75 \sqrt{41} - 68 \sqrt{41} - 48 \sqrt{41} - 64}{32}$

Subtract $64$ from $299$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{235 + 75 \sqrt{41} - 68 \sqrt{41} - 48 \sqrt{41}}{32}$

Subtract $68 \sqrt{41}$ from $75 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{235 + 7 \sqrt{41} - 48 \sqrt{41}}{32}$

Subtract $48 \sqrt{41}$ from $7 \sqrt{41}$ .

$f (\frac{3 - \sqrt{41}}{4}) = \frac{235 - 41 \sqrt{41}}{32}$

The final answer is $\frac{235 - 41 \sqrt{41}}{32}$ .

$y = \frac{235 - 41 \sqrt{41}}{32}$

These are the local extrema for $f (x) = - x^{4} + 2 x^{3} + 4 x^{2} - 2$ .

$(0, - 2)$ is a local minima

$(\frac{3 + \sqrt{41}}{4}, \frac{235 + 41 \sqrt{41}}{32})$ is a local maxima

$(\frac{3 - \sqrt{41}}{4}, \frac{235 - 41 \sqrt{41}}{32})$ is a local maxima