Calculus Examples

$f (x) = \frac{1}{4} x^{4} - x$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

Evaluate $\frac{d}{d x} [\frac{1}{4} x^{4}]$ .

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Combine $\frac{4}{4}$ and $x^{3}$ .

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

Step 1.2.5

Cancel the common factor of $4$ .

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Step 1.2.5.1

Cancel the common factor.

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

Step 1.2.5.2

Divide $x^{3}$ by $1$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

$x^{3} - 1 \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

$x^{3} - 1$

Step 2

Find the second derivative of the function.

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Step 2.1

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

$f'' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 1)$

Step 2.2

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

$f'' (x) = 3 x^{2} + \frac{d}{d x} (- 1)$

Step 2.3

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

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

Step 2.4

Add $3 x^{2}$ and $0$ .

$f'' (x) = 3 x^{2}$

Step 3

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

$x^{3} - 1 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

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

Step 4.1.2

Evaluate $\frac{d}{d x} [\frac{1}{4} x^{4}]$ .

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Step 4.1.2.1

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

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

Step 4.1.2.2

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

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

Step 4.1.2.3

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

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

Step 4.1.2.4

Combine $\frac{4}{4}$ and $x^{3}$ .

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

Step 4.1.2.5

Cancel the common factor of $4$ .

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Step 4.1.2.5.1

Cancel the common factor.

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

Step 4.1.2.5.2

Divide $x^{3}$ by $1$ .

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

Step 4.1.3

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

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Step 4.1.3.1

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

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

Step 4.1.3.2

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

$x^{3} - 1 \cdot 1$

Step 4.1.3.3

Multiply $- 1$ by $1$ .

$f' (x) = x^{3} - 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $x^{3} - 1$ .

$x^{3} - 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $x^{3} - 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$x^{3} - 1 = 0$

Step 5.2

Add $1$ to both sides of the equation.

$x^{3} = 1$

Step 5.3

Subtract $1$ from both sides of the equation.

$x^{3} - 1 = 0$

Step 5.4

Factor the left side of the equation.

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Step 5.4.1

Rewrite $1$ as $1^{3}$ .

$x^{3} - 1^{3} = 0$

Step 5.4.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Step 5.4.3

Simplify.

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Step 5.4.3.1

Multiply $x$ by $1$ .

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

Step 5.4.3.2

One to any power is one.

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

Step 5.5

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

$x - 1 = 0$

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

Step 5.6

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

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Step 5.6.1

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

$x - 1 = 0$

Step 5.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.7

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

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Step 5.7.1

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

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

Step 5.7.2

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

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Step 5.7.2.1

Use the quadratic formula to find the solutions.

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

Step 5.7.2.2

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

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

Step 5.7.2.3

Simplify.

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Step 5.7.2.3.1

Simplify the numerator.

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Step 5.7.2.3.1.1

One to any power is one.

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

Step 5.7.2.3.1.2

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

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Step 5.7.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.7.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 5.7.2.3.1.3

Subtract $4$ from $1$ .

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

Step 5.7.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 5.7.2.3.1.5

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

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

Step 5.7.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 5.7.2.3.2

Multiply $2$ by $1$ .

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

Step 5.7.2.4

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

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Step 5.7.2.4.1

Simplify the numerator.

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Step 5.7.2.4.1.1

One to any power is one.

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

Step 5.7.2.4.1.2

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

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Step 5.7.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.7.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 5.7.2.4.1.3

Subtract $4$ from $1$ .

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

Step 5.7.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 5.7.2.4.1.5

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

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

Step 5.7.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 5.7.2.4.2

Multiply $2$ by $1$ .

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

Step 5.7.2.4.3

Change the $\pm$ to $+$ .

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

Step 5.7.2.4.4

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

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

Step 5.7.2.4.5

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

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

Step 5.7.2.4.6

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

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

Step 5.7.2.4.7

Move the negative in front of the fraction.

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

Step 5.7.2.5

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

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Step 5.7.2.5.1

Simplify the numerator.

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Step 5.7.2.5.1.1

One to any power is one.

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

Step 5.7.2.5.1.2

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

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Step 5.7.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.7.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 5.7.2.5.1.3

Subtract $4$ from $1$ .

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

Step 5.7.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 5.7.2.5.1.5

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

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

Step 5.7.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

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

Step 5.7.2.5.2

Multiply $2$ by $1$ .

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

Step 5.7.2.5.3

Change the $\pm$ to $-$ .

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

Step 5.7.2.5.4

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

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

Step 5.7.2.5.5

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

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

Step 5.7.2.5.6

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

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

Step 5.7.2.5.7

Move the negative in front of the fraction.

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

Step 5.7.2.6

The final answer is the combination of both solutions.

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

Step 5.8

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

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

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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.

Step 7

Critical points to evaluate.

$x = 1$

Step 8

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$3 {(1)}^{2}$

Step 9

Evaluate the second derivative.

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Step 9.1

One to any power is one.

$3 \cdot 1$

Step 9.2

Multiply $3$ by $1$ .

$3$

Step 10

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 11

Find the y-value when $x = 1$ .

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Step 11.1

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

$f (1) = \frac{1}{4} \cdot {(1)}^{4} - (1)$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

One to any power is one.

$f (1) = \frac{1}{4} \cdot 1 - (1)$

Step 11.2.1.2

Multiply $\frac{1}{4}$ by $1$ .

$f (1) = \frac{1}{4} - (1)$

Step 11.2.1.3

Multiply $- 1$ by $1$ .

$f (1) = \frac{1}{4} - 1$

Step 11.2.2

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

$f (1) = \frac{1}{4} - 1 \cdot \frac{4}{4}$

Step 11.2.3

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

$f (1) = \frac{1}{4} + \frac{- 1 \cdot 4}{4}$

Step 11.2.4

Combine the numerators over the common denominator.

$f (1) = \frac{1 - 1 \cdot 4}{4}$

Step 11.2.5

Simplify the numerator.

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Step 11.2.5.1

Multiply $- 1$ by $4$ .

$f (1) = \frac{1 - 4}{4}$

Step 11.2.5.2

Subtract $4$ from $1$ .

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

Step 11.2.6

Move the negative in front of the fraction.

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

Step 11.2.7

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

$y = - \frac{3}{4}$

Step 12

These are the local extrema for $f (x) = \frac{1}{4} x^{4} - x$ .

$(1, - \frac{3}{4})$ is a local minima

Step 13