Calculus Examples

$y = x^{4} - 3 x^{2} + 2 x - 1$

Step 1

Set $y$ as a function of $x$ .

$f (x) = x^{4} - 3 x^{2} + 2 x - 1$

Step 2

Find the derivative.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{4} - 3 x^{2} + 2 x - 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.1.2

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

$4 x^{3} + \frac{d}{d x} [- 3 x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.2

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

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Step 2.2.1

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

$4 x^{3} - 3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.2.2

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

$4 x^{3} - 3 (2 x) + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.2.3

Multiply $2$ by $- 3$ .

$4 x^{3} - 6 x + \frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$

Step 2.3

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

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Step 2.3.1

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

$4 x^{3} - 6 x + 2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 2.3.2

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

$4 x^{3} - 6 x + 2 \cdot 1 + \frac{d}{d x} [- 1]$

Step 2.3.3

Multiply $2$ by $1$ .

$4 x^{3} - 6 x + 2 + \frac{d}{d x} [- 1]$

Step 2.4

Differentiate using the Constant Rule.

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Step 2.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 2.4.2

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

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

Step 3

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

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Step 3.1

Factor the left side of the equation.

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Step 3.1.1

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

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Step 3.1.1.1

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

$2 (2 x^{3}) - 6 x + 2 = 0$

Step 3.1.1.2

Factor $2$ out of $- 6 x$ .

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

Step 3.1.1.3

Factor $2$ out of $2$ .

$2 (2 x^{3}) + 2 (- 3 x) + 2 (1) = 0$

Step 3.1.1.4

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

$2 (2 x^{3} - 3 x) + 2 (1) = 0$

Step 3.1.1.5

Factor $2$ out of $2 (2 x^{3} - 3 x) + 2 (1)$ .

$2 (2 x^{3} - 3 x + 1) = 0$

Step 3.1.2

Factor.

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Step 3.1.2.1

Factor $2 x^{3} - 3 x + 1$ using the rational roots test.

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Step 3.1.2.1.1

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1$

$q = \pm 1, \pm 2$

Step 3.1.2.1.2

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.5$

Step 3.1.2.1.3

Substitute $1$ and simplify the expression. In this case, the expression is equal to $0$ so $1$ is a root of the polynomial.

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Step 3.1.2.1.3.1

Substitute $1$ into the polynomial.

$2 \cdot 1^{3} - 3 \cdot 1 + 1$

Step 3.1.2.1.3.2

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

$2 \cdot 1 - 3 \cdot 1 + 1$

Step 3.1.2.1.3.3

Multiply $2$ by $1$ .

$2 - 3 \cdot 1 + 1$

Step 3.1.2.1.3.4

Multiply $- 3$ by $1$ .

$2 - 3 + 1$

Step 3.1.2.1.3.5

Subtract $3$ from $2$ .

$- 1 + 1$

Step 3.1.2.1.3.6

Add $- 1$ and $1$ .

$0$

Step 3.1.2.1.4

Since $1$ is a known root, divide the polynomial by $x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{2 x^{3} - 3 x + 1}{x - 1}$

Step 3.1.2.1.5

Divide $2 x^{3} - 3 x + 1$ by $x - 1$ .

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Step 3.1.2.1.5.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$2 x^{3}$

$0 x^{2}$

$3 x$

$1$

Step 3.1.2.1.5.2

Divide the highest order term in the dividend $2 x^{3}$ by the highest order term in divisor $x$ .

					$2 x^{2}$
	$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$

Step 3.1.2.1.5.3

Multiply the new quotient term by the divisor.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			+	$2 x^{3}$	-	$2 x^{2}$

Step 3.1.2.1.5.4

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{3} - 2 x^{2}$

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$

Step 3.1.2.1.5.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$

Step 3.1.2.1.5.6

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$

Step 3.1.2.1.5.7

Divide the highest order term in the dividend $2 x^{2}$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$2 x$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$

Step 3.1.2.1.5.8

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$2 x$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					+	$2 x^{2}$	-	$2 x$

Step 3.1.2.1.5.9

The expression needs to be subtracted from the dividend, so change all the signs in $2 x^{2} - 2 x$

				$2 x^{2}$	+	$2 x$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$

Step 3.1.2.1.5.10

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	+	$2 x$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$

Step 3.1.2.1.5.11

Pull the next terms from the original dividend down into the current dividend.

				$2 x^{2}$	+	$2 x$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$	+	$1$

Step 3.1.2.1.5.12

Divide the highest order term in the dividend $- x$ by the highest order term in divisor $x$ .

				$2 x^{2}$	+	$2 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$	+	$1$

Step 3.1.2.1.5.13

Multiply the new quotient term by the divisor.

				$2 x^{2}$	+	$2 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$	+	$1$
							-	$x$	+	$1$

Step 3.1.2.1.5.14

The expression needs to be subtracted from the dividend, so change all the signs in $- x + 1$

				$2 x^{2}$	+	$2 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$	+	$1$
							+	$x$	-	$1$

Step 3.1.2.1.5.15

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$2 x^{2}$	+	$2 x$	-	$1$
$x$	-	$1$		$2 x^{3}$	+	$0 x^{2}$	-	$3 x$	+	$1$
			-	$2 x^{3}$	+	$2 x^{2}$
					+	$2 x^{2}$	-	$3 x$
					-	$2 x^{2}$	+	$2 x$
							-	$x$	+	$1$
							+	$x$	-	$1$
										$0$

Step 3.1.2.1.5.16

Since the remander is $0$ , the final answer is the quotient.

$2 x^{2} + 2 x - 1$

Step 3.1.2.1.6

Write $2 x^{3} - 3 x + 1$ as a set of factors.

$2 ((x - 1) (2 x^{2} + 2 x - 1)) = 0$

Step 3.1.2.2

Remove unnecessary parentheses.

$2 (x - 1) (2 x^{2} + 2 x - 1) = 0$

Step 3.2

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

$x - 1 = 0$

$2 x^{2} + 2 x - 1 = 0$

Step 3.3

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 3.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 3.4

Set $2 x^{2} + 2 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $2 x^{2} + 2 x - 1$ equal to $0$ .

$2 x^{2} + 2 x - 1 = 0$

Step 3.4.2

Solve $2 x^{2} + 2 x - 1 = 0$ for $x$ .

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Step 3.4.2.1

Use the quadratic formula to find the solutions.

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

Step 3.4.2.2

Substitute the values $a = 2$ , $b = 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (2 \cdot - 1)}}{2 \cdot 2}$

Step 3.4.2.3

Simplify.

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Step 3.4.2.3.1

Simplify the numerator.

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Step 3.4.2.3.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.3.1.2

Multiply $- 4 \cdot 2 \cdot - 1$ .

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Step 3.4.2.3.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.3.1.2.2

Multiply $- 8$ by $- 1$ .

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

Step 3.4.2.3.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 2}$

Step 3.4.2.3.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 3.4.2.3.1.4.1

Factor $4$ out of $12$ .

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

Step 3.4.2.3.1.4.2

Rewrite $4$ as $2^{2}$ .

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

Step 3.4.2.3.1.5

Pull terms out from under the radical.

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

Step 3.4.2.3.2

Multiply $2$ by $2$ .

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

Step 3.4.2.3.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{4}$ .

$x = \frac{- 1 \pm \sqrt{3}}{2}$

Step 3.4.2.4

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

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Step 3.4.2.4.1

Simplify the numerator.

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Step 3.4.2.4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.4.1.2

Multiply $- 4 \cdot 2 \cdot - 1$ .

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Step 3.4.2.4.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.4.1.2.2

Multiply $- 8$ by $- 1$ .

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

Step 3.4.2.4.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 2}$

Step 3.4.2.4.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 3.4.2.4.1.4.1

Factor $4$ out of $12$ .

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

Step 3.4.2.4.1.4.2

Rewrite $4$ as $2^{2}$ .

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

Step 3.4.2.4.1.5

Pull terms out from under the radical.

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

Step 3.4.2.4.2

Multiply $2$ by $2$ .

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

Step 3.4.2.4.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{4}$ .

$x = \frac{- 1 \pm \sqrt{3}}{2}$

Step 3.4.2.4.4

Change the $\pm$ to $+$ .

$x = \frac{- 1 + \sqrt{3}}{2}$

Step 3.4.2.4.5

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

$x = \frac{- 1 \cdot 1 + \sqrt{3}}{2}$

Step 3.4.2.4.6

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

$x = \frac{- 1 \cdot 1 - 1 (- \sqrt{3})}{2}$

Step 3.4.2.4.7

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

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

Step 3.4.2.4.8

Move the negative in front of the fraction.

$x = - \frac{1 - \sqrt{3}}{2}$

Step 3.4.2.5

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

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Step 3.4.2.5.1

Simplify the numerator.

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Step 3.4.2.5.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 2 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.5.1.2

Multiply $- 4 \cdot 2 \cdot - 1$ .

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Step 3.4.2.5.1.2.1

Multiply $- 4$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 8 \cdot - 1}}{2 \cdot 2}$

Step 3.4.2.5.1.2.2

Multiply $- 8$ by $- 1$ .

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

Step 3.4.2.5.1.3

Add $4$ and $8$ .

$x = \frac{- 2 \pm \sqrt{12}}{2 \cdot 2}$

Step 3.4.2.5.1.4

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 3.4.2.5.1.4.1

Factor $4$ out of $12$ .

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

Step 3.4.2.5.1.4.2

Rewrite $4$ as $2^{2}$ .

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

Step 3.4.2.5.1.5

Pull terms out from under the radical.

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

Step 3.4.2.5.2

Multiply $2$ by $2$ .

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

Step 3.4.2.5.3

Simplify $\frac{- 2 \pm 2 \sqrt{3}}{4}$ .

$x = \frac{- 1 \pm \sqrt{3}}{2}$

Step 3.4.2.5.4

Change the $\pm$ to $-$ .

$x = \frac{- 1 - \sqrt{3}}{2}$

Step 3.4.2.5.5

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

$x = \frac{- 1 \cdot 1 - \sqrt{3}}{2}$

Step 3.4.2.5.6

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

$x = \frac{- 1 \cdot 1 - (\sqrt{3})}{2}$

Step 3.4.2.5.7

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

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

Step 3.4.2.5.8

Move the negative in front of the fraction.

$x = - \frac{1 + \sqrt{3}}{2}$

Step 3.4.2.6

The final answer is the combination of both solutions.

$x = - \frac{1 - \sqrt{3}}{2}, - \frac{1 + \sqrt{3}}{2}$

Step 3.5

The final solution is all the values that make $2 (x - 1) (2 x^{2} + 2 x - 1) = 0$ true.

$x = 1, - \frac{1 - \sqrt{3}}{2}, - \frac{1 + \sqrt{3}}{2}$

Step 4

Solve the original function $f (x) = x^{4} - 3 x^{2} + 2 x - 1$ at $x = 1$ .

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Step 4.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{4} - 3 {(1)}^{2} + 2 (1) - 1$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

$f (1) = 1 - 3 {(1)}^{2} + 2 (1) - 1$

Step 4.2.1.2

One to any power is one.

$f (1) = 1 - 3 \cdot 1 + 2 (1) - 1$

Step 4.2.1.3

Multiply $- 3$ by $1$ .

$f (1) = 1 - 3 + 2 (1) - 1$

Step 4.2.1.4

Multiply $2$ by $1$ .

$f (1) = 1 - 3 + 2 - 1$

Step 4.2.2

Simplify by adding and subtracting.

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Step 4.2.2.1

Subtract $3$ from $1$ .

$f (1) = - 2 + 2 - 1$

Step 4.2.2.2

Add $- 2$ and $2$ .

$f (1) = 0 - 1$

Step 4.2.2.3

Subtract $1$ from $0$ .

$f (1) = - 1$

Step 4.2.3

The final answer is $- 1$ .

$- 1$

Step 5

Solve the original function $f (x) = x^{4} - 3 x^{2} + 2 x - 1$ at $x = - \frac{1 - \sqrt{3}}{2}$ .

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Step 5.1

Replace the variable $x$ with $- \frac{1 - \sqrt{3}}{2}$ in the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = {(- \frac{1 - \sqrt{3}}{2})}^{4} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 5.2.1.1.1

Apply the product rule to $- \frac{1 - \sqrt{3}}{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = {(- 1)}^{4} {(\frac{1 - \sqrt{3}}{2})}^{4} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.1.2

Apply the product rule to $\frac{1 - \sqrt{3}}{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = {(- 1)}^{4} (\frac{{(1 - \sqrt{3})}^{4}}{2^{4}}) - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.2

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

$f (- \frac{1 - \sqrt{3}}{2}) = 1 (\frac{{(1 - \sqrt{3})}^{4}}{2^{4}}) - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.3

Multiply $\frac{{(1 - \sqrt{3})}^{4}}{2^{4}}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{{(1 - \sqrt{3})}^{4}}{2^{4}} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.4

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{{(1 - \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.5

Use the Binomial Theorem.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1^{4} + 4 \cdot (1^{3} (- \sqrt{3})) + 6 \cdot (1^{2} {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6

Simplify each term.

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Step 5.2.1.6.1

One to any power is one.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 + 4 \cdot (1^{3} (- \sqrt{3})) + 6 \cdot (1^{2} {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.2

One to any power is one.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 + 4 \cdot (1 (- \sqrt{3})) + 6 \cdot (1^{2} {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.3

Multiply $4$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 + 4 (- \sqrt{3}) + 6 \cdot (1^{2} {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.4

Multiply $- 1$ by $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot (1^{2} {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.5

One to any power is one.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot (1 {(- \sqrt{3})}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.6

Multiply $6$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 {(- \sqrt{3})}^{2} + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.7

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 ({(- 1)}^{2} {\sqrt{3}}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.8

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 (1 {\sqrt{3}}^{2}) + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.9

Multiply ${\sqrt{3}}^{2}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 {\sqrt{3}}^{2} + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.10

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 5.2.1.6.10.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 {(3^{\frac{1}{2}})}^{2} + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.10.2

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot 3^{\frac{1}{2} \cdot 2} + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.10.3

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot 3^{\frac{2}{2}} + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.10.4

Cancel the common factor of $2$ .

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Step 5.2.1.6.10.4.1

Cancel the common factor.

Step 5.2.1.6.10.4.2

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot 3 + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.10.5

Evaluate the exponent.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 6 \cdot 3 + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.11

Multiply $6$ by $3$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 \cdot (1 {(- \sqrt{3})}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.12

Multiply $4$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 {(- \sqrt{3})}^{3} + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.13

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 ({(- 1)}^{3} {\sqrt{3}}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.14

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- {\sqrt{3}}^{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.15

Rewrite ${\sqrt{3}}^{3}$ as $\sqrt{3^{3}}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- \sqrt{3^{3}}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.16

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- \sqrt{27}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.17

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 5.2.1.6.17.1

Factor $9$ out of $27$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- \sqrt{9 (3)}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.17.2

Rewrite $9$ as $3^{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- \sqrt{3^{2} \cdot 3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.18

Pull terms out from under the radical.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- (3 \sqrt{3})) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.19

Multiply $3$ by $- 1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 + 4 (- 3 \sqrt{3}) + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.20

Multiply $- 3$ by $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + {(- \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.21

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + {(- 1)}^{4} {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.22

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 1 {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.23

Multiply ${\sqrt{3}}^{4}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

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Step 5.2.1.6.24.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + {(3^{\frac{1}{2}})}^{4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.2

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{1}{2} \cdot 4}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.3

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{4}{2}}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.4

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

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Step 5.2.1.6.24.4.1

Factor $2$ out of $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2}}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.4.2

Cancel the common factors.

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Step 5.2.1.6.24.4.2.1

Factor $2$ out of $2$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2 (1)}}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.4.2.2

Cancel the common factor.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2 \cdot 1}}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.4.2.3

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{\frac{2}{1}}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.24.4.2.4

Divide $2$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 3^{2}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.6.25

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{1 - 4 \sqrt{3} + 18 - 12 \sqrt{3} + 9}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.7

Add $1$ and $18$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{19 - 4 \sqrt{3} - 12 \sqrt{3} + 9}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.8

Add $19$ and $9$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{28 - 4 \sqrt{3} - 12 \sqrt{3}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.9

Subtract $12 \sqrt{3}$ from $- 4 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{28 - 16 \sqrt{3}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10

Cancel the common factor of $28 - 16 \sqrt{3}$ and $16$ .

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Step 5.2.1.10.1

Factor $4$ out of $28$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{4 (7) - 16 \sqrt{3}}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10.2

Factor $4$ out of $- 16 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{4 (7) + 4 (- 4 \sqrt{3})}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10.3

Factor $4$ out of $4 (7) + 4 (- 4 \sqrt{3})$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{4 (7 - 4 \sqrt{3})}{16} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10.4

Cancel the common factors.

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Step 5.2.1.10.4.1

Factor $4$ out of $16$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{4 (7 - 4 \sqrt{3})}{4 \cdot 4} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10.4.2

Cancel the common factor.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{4 (7 - 4 \sqrt{3})}{4 \cdot 4} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.10.4.3

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 {(- \frac{1 - \sqrt{3}}{2})}^{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.11

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 5.2.1.11.1

Apply the product rule to $- \frac{1 - \sqrt{3}}{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 ({(- 1)}^{2} {(\frac{1 - \sqrt{3}}{2})}^{2}) + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.11.2

Apply the product rule to $\frac{1 - \sqrt{3}}{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 ({(- 1)}^{2} (\frac{{(1 - \sqrt{3})}^{2}}{2^{2}})) + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.12

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 (1 (\frac{{(1 - \sqrt{3})}^{2}}{2^{2}})) + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.13

Multiply $\frac{{(1 - \sqrt{3})}^{2}}{2^{2}}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{{(1 - \sqrt{3})}^{2}}{2^{2}} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.14

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{{(1 - \sqrt{3})}^{2}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.15

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{(1 - \sqrt{3}) (1 - \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.16

Expand $(1 - \sqrt{3}) (1 - \sqrt{3})$ using the FOIL Method.

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Step 5.2.1.16.1

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 (1 - \sqrt{3}) - \sqrt{3} (1 - \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.16.2

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 \cdot 1 + 1 (- \sqrt{3}) - \sqrt{3} (1 - \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.16.3

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 \cdot 1 + 1 (- \sqrt{3}) - \sqrt{3} \cdot 1 - \sqrt{3} (- \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17

Simplify and combine like terms.

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Step 5.2.1.17.1

Simplify each term.

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Step 5.2.1.17.1.1

Multiply $1$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 + 1 (- \sqrt{3}) - \sqrt{3} \cdot 1 - \sqrt{3} (- \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.2

Multiply $- \sqrt{3}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} \cdot 1 - \sqrt{3} (- \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.3

Multiply $- 1$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} - \sqrt{3} (- \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4

Multiply $- \sqrt{3} (- \sqrt{3})$ .

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Step 5.2.1.17.1.4.1

Multiply $- 1$ by $- 1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 1 \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4.2

Multiply $\sqrt{3}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4.3

Raise $\sqrt{3}$ to the power of $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4.4

Raise $\sqrt{3}$ to the power of $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4.5

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + {\sqrt{3}}^{1 + 1}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.4.6

Add $1$ and $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + {\sqrt{3}}^{2}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 5.2.1.17.1.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + {(3^{\frac{1}{2}})}^{2}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5.2

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 3^{\frac{1}{2} \cdot 2}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5.3

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 3^{\frac{2}{2}}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5.4

Cancel the common factor of $2$ .

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Step 5.2.1.17.1.5.4.1

Cancel the common factor.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 3^{\frac{2}{2}}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5.4.2

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 3}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.1.5.5

Evaluate the exponent.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{1 - \sqrt{3} - \sqrt{3} + 3}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.2

Add $1$ and $3$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{4 - \sqrt{3} - \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.17.3

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{4 - 2 \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18

Cancel the common factor of $4 - 2 \sqrt{3}$ and $4$ .

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Step 5.2.1.18.1

Factor $2$ out of $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 (2) - 2 \sqrt{3}}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18.2

Factor $2$ out of $- 2 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 (2) + 2 (- \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18.3

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 (2 - \sqrt{3})}{4} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18.4

Cancel the common factors.

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Step 5.2.1.18.4.1

Factor $2$ out of $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 (2 - \sqrt{3})}{2 \cdot 2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18.4.2

Cancel the common factor.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 (2 - \sqrt{3})}{2 \cdot 2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.18.4.3

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - 3 \frac{2 - \sqrt{3}}{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.19

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} + \frac{- 3 (2 - \sqrt{3})}{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.20

Move the negative in front of the fraction.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{(3) (2 - \sqrt{3})}{2} + 2 (- \frac{1 - \sqrt{3}}{2}) - 1$

Step 5.2.1.21

Cancel the common factor of $2$ .

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Step 5.2.1.21.1

Move the leading negative in $- \frac{1 - \sqrt{3}}{2}$ into the numerator.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} + 2 (\frac{- (1 - \sqrt{3})}{2}) - 1$

Step 5.2.1.21.2

Cancel the common factor.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} + 2 (\frac{- (1 - \sqrt{3})}{2}) - 1$

Step 5.2.1.21.3

Rewrite the expression.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} - (1 - \sqrt{3}) - 1$

Step 5.2.1.22

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} - 1 \cdot 1 + \sqrt{3} - 1$

Step 5.2.1.23

Multiply $- 1$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} - 1 + \sqrt{3} - 1$

Step 5.2.1.24

Multiply $- - \sqrt{3}$ .

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Step 5.2.1.24.1

Multiply $- 1$ by $- 1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} - 1 + 1 \sqrt{3} - 1$

Step 5.2.1.24.2

Multiply $\sqrt{3}$ by $1$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} - 1 + \sqrt{3} - 1$

Step 5.2.2

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3})}{2} \cdot \frac{2}{2} - 1 + \sqrt{3} - 1$

Step 5.2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.3.1

Multiply $\frac{3 (2 - \sqrt{3})}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3}) \cdot 2}{2 \cdot 2} - 1 + \sqrt{3} - 1$

Step 5.2.3.2

Multiply $2$ by $2$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3}}{4} - \frac{3 (2 - \sqrt{3}) \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.4

Combine the numerators over the common denominator.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} - 3 (2 - \sqrt{3}) \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5

Simplify the numerator.

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Step 5.2.5.1

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} + (- 3 \cdot 2 - 3 (- \sqrt{3})) \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.2

Multiply $- 3$ by $2$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} + (- 6 - 3 (- \sqrt{3})) \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.3

Multiply $- 1$ by $- 3$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} + (- 6 + 3 \sqrt{3}) \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.4

Apply the distributive property.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} - 6 \cdot 2 + 3 \sqrt{3} \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.5

Multiply $- 6$ by $2$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} - 12 + 3 \sqrt{3} \cdot 2}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.6

Multiply $2$ by $3$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{7 - 4 \sqrt{3} - 12 + 6 \sqrt{3}}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.7

Subtract $12$ from $7$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 - 4 \sqrt{3} + 6 \sqrt{3}}{4} - 1 + \sqrt{3} - 1$

Step 5.2.5.8

Add $- 4 \sqrt{3}$ and $6 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 + 2 \sqrt{3}}{4} - 1 + \sqrt{3} - 1$

Step 5.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 + 2 \sqrt{3}}{4} - 1 \cdot \frac{4}{4} + \sqrt{3} - 1$

Step 5.2.7

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 + 2 \sqrt{3}}{4} + \frac{- 1 \cdot 4}{4} + \sqrt{3} - 1$

Step 5.2.8

Simplify the expression.

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Step 5.2.8.1

Combine the numerators over the common denominator.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 + 2 \sqrt{3} - 1 \cdot 4}{4} + \sqrt{3} - 1$

Step 5.2.8.2

Multiply $- 1$ by $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 5 + 2 \sqrt{3} - 4}{4} + \sqrt{3} - 1$

Step 5.2.8.3

Subtract $4$ from $- 5$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 2 \sqrt{3}}{4} + \sqrt{3} - 1$

Step 5.2.9

To write $\sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 2 \sqrt{3}}{4} + \sqrt{3} \cdot \frac{4}{4} - 1$

Step 5.2.10

Combine $\sqrt{3}$ and $\frac{4}{4}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 2 \sqrt{3}}{4} + \frac{\sqrt{3} \cdot 4}{4} - 1$

Step 5.2.11

Simplify the expression.

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Step 5.2.11.1

Combine the numerators over the common denominator.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 2 \sqrt{3} + \sqrt{3} \cdot 4}{4} - 1$

Step 5.2.11.2

Reorder the factors of $\sqrt{3} \cdot 4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 2 \sqrt{3} + 4 \sqrt{3}}{4} - 1$

Step 5.2.12

Add $2 \sqrt{3}$ and $4 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 6 \sqrt{3}}{4} - 1$

Step 5.2.13

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 6 \sqrt{3}}{4} - 1 \cdot \frac{4}{4}$

Step 5.2.14

Combine fractions.

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Step 5.2.14.1

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

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 6 \sqrt{3}}{4} + \frac{- 1 \cdot 4}{4}$

Step 5.2.14.2

Combine the numerators over the common denominator.

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 6 \sqrt{3} - 1 \cdot 4}{4}$

Step 5.2.15

Simplify the numerator.

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Step 5.2.15.1

Multiply $- 1$ by $4$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 9 + 6 \sqrt{3} - 4}{4}$

Step 5.2.15.2

Subtract $4$ from $- 9$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 13 + 6 \sqrt{3}}{4}$

Step 5.2.16

Simplify with factoring out.

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Step 5.2.16.1

Rewrite $- 13$ as $- 1 (13)$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 1 \cdot 13 + 6 \sqrt{3}}{4}$

Step 5.2.16.2

Factor $- 1$ out of $6 \sqrt{3}$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 1 \cdot 13 - (- 6 \sqrt{3})}{4}$

Step 5.2.16.3

Factor $- 1$ out of $- 1 (13) - (- 6 \sqrt{3})$ .

$f (- \frac{1 - \sqrt{3}}{2}) = \frac{- 1 (13 - 6 \sqrt{3})}{4}$

Step 5.2.16.4

Move the negative in front of the fraction.

$f (- \frac{1 - \sqrt{3}}{2}) = - \frac{13 - 6 \sqrt{3}}{4}$

Step 5.2.17

The final answer is $- \frac{13 - 6 \sqrt{3}}{4}$ .

$- \frac{13 - 6 \sqrt{3}}{4}$

Step 6

Solve the original function $f (x) = x^{4} - 3 x^{2} + 2 x - 1$ at $x = - \frac{1 + \sqrt{3}}{2}$ .

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Step 6.1

Replace the variable $x$ with $- \frac{1 + \sqrt{3}}{2}$ in the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = {(- \frac{1 + \sqrt{3}}{2})}^{4} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.2.1.1.1

Apply the product rule to $- \frac{1 + \sqrt{3}}{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = {(- 1)}^{4} {(\frac{1 + \sqrt{3}}{2})}^{4} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.1.2

Apply the product rule to $\frac{1 + \sqrt{3}}{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = {(- 1)}^{4} (\frac{{(1 + \sqrt{3})}^{4}}{2^{4}}) - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.2

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

$f (- \frac{1 + \sqrt{3}}{2}) = 1 (\frac{{(1 + \sqrt{3})}^{4}}{2^{4}}) - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.3

Multiply $\frac{{(1 + \sqrt{3})}^{4}}{2^{4}}$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{{(1 + \sqrt{3})}^{4}}{2^{4}} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.4

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{{(1 + \sqrt{3})}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.5

Use the Binomial Theorem.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1^{4} + 4 \cdot (1^{3} \sqrt{3}) + 6 \cdot (1^{2} {\sqrt{3}}^{2}) + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6

Simplify each term.

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Step 6.2.1.6.1

One to any power is one.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \cdot (1^{3} \sqrt{3}) + 6 \cdot (1^{2} {\sqrt{3}}^{2}) + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.2

One to any power is one.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \cdot (1 \sqrt{3}) + 6 \cdot (1^{2} {\sqrt{3}}^{2}) + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.3

Multiply $4$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot (1^{2} {\sqrt{3}}^{2}) + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.4

One to any power is one.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot (1 {\sqrt{3}}^{2}) + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.5

Multiply $6$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 {\sqrt{3}}^{2} + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.6

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 6.2.1.6.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 {(3^{\frac{1}{2}})}^{2} + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.6.2

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot 3^{\frac{1}{2} \cdot 2} + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.6.3

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot 3^{\frac{2}{2}} + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.6.4

Cancel the common factor of $2$ .

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Step 6.2.1.6.6.4.1

Cancel the common factor.

Step 6.2.1.6.6.4.2

Rewrite the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot 3 + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.6.5

Evaluate the exponent.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 6 \cdot 3 + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.7

Multiply $6$ by $3$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 \cdot (1 {\sqrt{3}}^{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.8

Multiply $4$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 {\sqrt{3}}^{3} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.9

Rewrite ${\sqrt{3}}^{3}$ as $\sqrt{3^{3}}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 \sqrt{3^{3}} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.10

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 \sqrt{27} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.11

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 6.2.1.6.11.1

Factor $9$ out of $27$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 \sqrt{9 (3)} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.11.2

Rewrite $9$ as $3^{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 \sqrt{3^{2} \cdot 3} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.12

Pull terms out from under the radical.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 4 (3 \sqrt{3}) + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.13

Multiply $3$ by $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + {\sqrt{3}}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14

Rewrite ${\sqrt{3}}^{4}$ as $3^{2}$ .

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Step 6.2.1.6.14.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + {(3^{\frac{1}{2}})}^{4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.2

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{1}{2} \cdot 4}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.3

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{4}{2}}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.4

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

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Step 6.2.1.6.14.4.1

Factor $2$ out of $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2}}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.4.2

Cancel the common factors.

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Step 6.2.1.6.14.4.2.1

Factor $2$ out of $2$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2 (1)}}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.4.2.2

Cancel the common factor.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{2 \cdot 2}{2 \cdot 1}}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.4.2.3

Rewrite the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{\frac{2}{1}}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.14.4.2.4

Divide $2$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 3^{2}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.6.15

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{1 + 4 \sqrt{3} + 18 + 12 \sqrt{3} + 9}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.7

Add $1$ and $18$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{19 + 4 \sqrt{3} + 12 \sqrt{3} + 9}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.8

Add $19$ and $9$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{28 + 4 \sqrt{3} + 12 \sqrt{3}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.9

Add $4 \sqrt{3}$ and $12 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{28 + 16 \sqrt{3}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10

Cancel the common factor of $28 + 16 \sqrt{3}$ and $16$ .

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Step 6.2.1.10.1

Factor $4$ out of $28$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{4 (7) + 16 \sqrt{3}}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10.2

Factor $4$ out of $16 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{4 (7) + 4 (4 \sqrt{3})}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10.3

Factor $4$ out of $4 (7) + 4 (4 \sqrt{3})$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{4 (7 + 4 \sqrt{3})}{16} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10.4

Cancel the common factors.

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Step 6.2.1.10.4.1

Factor $4$ out of $16$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{4 (7 + 4 \sqrt{3})}{4 \cdot 4} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10.4.2

Cancel the common factor.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{4 (7 + 4 \sqrt{3})}{4 \cdot 4} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.10.4.3

Rewrite the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 {(- \frac{1 + \sqrt{3}}{2})}^{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.11

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 6.2.1.11.1

Apply the product rule to $- \frac{1 + \sqrt{3}}{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 ({(- 1)}^{2} {(\frac{1 + \sqrt{3}}{2})}^{2}) + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.11.2

Apply the product rule to $\frac{1 + \sqrt{3}}{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 ({(- 1)}^{2} (\frac{{(1 + \sqrt{3})}^{2}}{2^{2}})) + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.12

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 (1 (\frac{{(1 + \sqrt{3})}^{2}}{2^{2}})) + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.13

Multiply $\frac{{(1 + \sqrt{3})}^{2}}{2^{2}}$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{{(1 + \sqrt{3})}^{2}}{2^{2}} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.14

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{{(1 + \sqrt{3})}^{2}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.15

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{(1 + \sqrt{3}) (1 + \sqrt{3})}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.16

Expand $(1 + \sqrt{3}) (1 + \sqrt{3})$ using the FOIL Method.

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Step 6.2.1.16.1

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 (1 + \sqrt{3}) + \sqrt{3} (1 + \sqrt{3})}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.16.2

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 \cdot 1 + 1 \sqrt{3} + \sqrt{3} (1 + \sqrt{3})}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.16.3

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 \cdot 1 + 1 \sqrt{3} + \sqrt{3} \cdot 1 + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17

Simplify and combine like terms.

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Step 6.2.1.17.1

Simplify each term.

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Step 6.2.1.17.1.1

Multiply $1$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + 1 \sqrt{3} + \sqrt{3} \cdot 1 + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.2

Multiply $\sqrt{3}$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} \cdot 1 + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.3

Multiply $\sqrt{3}$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} + \sqrt{3} \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.4

Combine using the product rule for radicals.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} + \sqrt{3 \cdot 3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.5

Multiply $3$ by $3$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} + \sqrt{9}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.6

Rewrite $9$ as $3^{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} + \sqrt{3^{2}}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.1.7

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{1 + \sqrt{3} + \sqrt{3} + 3}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.2

Add $1$ and $3$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{4 + \sqrt{3} + \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.17.3

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{4 + 2 \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18

Cancel the common factor of $4 + 2 \sqrt{3}$ and $4$ .

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Step 6.2.1.18.1

Factor $2$ out of $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 \cdot 2 + 2 \sqrt{3}}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18.2

Factor $2$ out of $2 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 \cdot 2 + 2 (\sqrt{3})}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18.3

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 \cdot (2 + \sqrt{3})}{4} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18.4

Cancel the common factors.

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Step 6.2.1.18.4.1

Factor $2$ out of $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 \cdot (2 + \sqrt{3})}{2 (2)} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18.4.2

Cancel the common factor.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 \cdot (2 + \sqrt{3})}{2 \cdot 2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.18.4.3

Rewrite the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - 3 \frac{2 + \sqrt{3}}{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.19

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} + \frac{- 3 (2 + \sqrt{3})}{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.20

Move the negative in front of the fraction.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{(3) (2 + \sqrt{3})}{2} + 2 (- \frac{1 + \sqrt{3}}{2}) - 1$

Step 6.2.1.21

Cancel the common factor of $2$ .

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Step 6.2.1.21.1

Move the leading negative in $- \frac{1 + \sqrt{3}}{2}$ into the numerator.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} + 2 (\frac{- (1 + \sqrt{3})}{2}) - 1$

Step 6.2.1.21.2

Cancel the common factor.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} + 2 (\frac{- (1 + \sqrt{3})}{2}) - 1$

Step 6.2.1.21.3

Rewrite the expression.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} - (1 + \sqrt{3}) - 1$

Step 6.2.1.22

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} - 1 \cdot 1 - \sqrt{3} - 1$

Step 6.2.1.23

Multiply $- 1$ by $1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} - 1 - \sqrt{3} - 1$

Step 6.2.2

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3})}{2} \cdot \frac{2}{2} - 1 - \sqrt{3} - 1$

Step 6.2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 6.2.3.1

Multiply $\frac{3 (2 + \sqrt{3})}{2}$ by $\frac{2}{2}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3}) \cdot 2}{2 \cdot 2} - 1 - \sqrt{3} - 1$

Step 6.2.3.2

Multiply $2$ by $2$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3}}{4} - \frac{3 (2 + \sqrt{3}) \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.4

Combine the numerators over the common denominator.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} - 3 (2 + \sqrt{3}) \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5

Simplify the numerator.

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Step 6.2.5.1

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} + (- 3 \cdot 2 - 3 \sqrt{3}) \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.2

Multiply $- 3$ by $2$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} + (- 6 - 3 \sqrt{3}) \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.3

Apply the distributive property.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} - 6 \cdot 2 - 3 \sqrt{3} \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.4

Multiply $- 6$ by $2$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} - 12 - 3 \sqrt{3} \cdot 2}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.5

Multiply $2$ by $- 3$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{7 + 4 \sqrt{3} - 12 - 6 \sqrt{3}}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.6

Subtract $12$ from $7$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 + 4 \sqrt{3} - 6 \sqrt{3}}{4} - 1 - \sqrt{3} - 1$

Step 6.2.5.7

Subtract $6 \sqrt{3}$ from $4 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 - 2 \sqrt{3}}{4} - 1 - \sqrt{3} - 1$

Step 6.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 - 2 \sqrt{3}}{4} - 1 \cdot \frac{4}{4} - \sqrt{3} - 1$

Step 6.2.7

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 - 2 \sqrt{3}}{4} + \frac{- 1 \cdot 4}{4} - \sqrt{3} - 1$

Step 6.2.8

Simplify the expression.

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Step 6.2.8.1

Combine the numerators over the common denominator.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 - 2 \sqrt{3} - 1 \cdot 4}{4} - \sqrt{3} - 1$

Step 6.2.8.2

Multiply $- 1$ by $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 5 - 2 \sqrt{3} - 4}{4} - \sqrt{3} - 1$

Step 6.2.8.3

Subtract $4$ from $- 5$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 2 \sqrt{3}}{4} - \sqrt{3} - 1$

Step 6.2.9

To write $- \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 2 \sqrt{3}}{4} - \sqrt{3} \cdot \frac{4}{4} - 1$

Step 6.2.10

Combine fractions.

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Step 6.2.10.1

Combine $- \sqrt{3}$ and $\frac{4}{4}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 2 \sqrt{3}}{4} + \frac{- \sqrt{3} \cdot 4}{4} - 1$

Step 6.2.10.2

Combine the numerators over the common denominator.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 2 \sqrt{3} - \sqrt{3} \cdot 4}{4} - 1$

Step 6.2.11

Simplify the numerator.

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Step 6.2.11.1

Multiply $4$ by $- 1$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 2 \sqrt{3} - 4 \sqrt{3}}{4} - 1$

Step 6.2.11.2

Subtract $4 \sqrt{3}$ from $- 2 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 6 \sqrt{3}}{4} - 1$

Step 6.2.12

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 6 \sqrt{3}}{4} - 1 \cdot \frac{4}{4}$

Step 6.2.13

Combine fractions.

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Step 6.2.13.1

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

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 6 \sqrt{3}}{4} + \frac{- 1 \cdot 4}{4}$

Step 6.2.13.2

Combine the numerators over the common denominator.

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 6 \sqrt{3} - 1 \cdot 4}{4}$

Step 6.2.14

Simplify the numerator.

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Step 6.2.14.1

Multiply $- 1$ by $4$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 9 - 6 \sqrt{3} - 4}{4}$

Step 6.2.14.2

Subtract $4$ from $- 9$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 13 - 6 \sqrt{3}}{4}$

Step 6.2.15

Simplify with factoring out.

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Step 6.2.15.1

Rewrite $- 13$ as $- 1 (13)$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 1 \cdot 13 - 6 \sqrt{3}}{4}$

Step 6.2.15.2

Factor $- 1$ out of $- 6 \sqrt{3}$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 1 \cdot 13 - (6 \sqrt{3})}{4}$

Step 6.2.15.3

Factor $- 1$ out of $- 1 (13) - (6 \sqrt{3})$ .

$f (- \frac{1 + \sqrt{3}}{2}) = \frac{- 1 (13 + 6 \sqrt{3})}{4}$

Step 6.2.15.4

Move the negative in front of the fraction.

$f (- \frac{1 + \sqrt{3}}{2}) = - \frac{13 + 6 \sqrt{3}}{4}$

Step 6.2.16

The final answer is $- \frac{13 + 6 \sqrt{3}}{4}$ .

$- \frac{13 + 6 \sqrt{3}}{4}$

Step 7

The horizontal tangent lines on function $f (x) = x^{4} - 3 x^{2} + 2 x - 1$ are $y = - 1, y = - \frac{13 - 6 \sqrt{3}}{4}, y = - \frac{13 + 6 \sqrt{3}}{4}$ .

$y = - 1, y = - \frac{13 - 6 \sqrt{3}}{4}, y = - \frac{13 + 6 \sqrt{3}}{4}$

Step 8