Calculus Examples

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

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of

x^{4} - x^{2} + x

$x^{4} - x^{2} + x$ with respect to

x

$x$ is

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

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

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

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

Step 1.1.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 4

$n = 4$ .

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

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

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

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

Step 1.2

Evaluate

\frac{d}{d x} [- x^{2}]

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

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Step 1.2.1

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- x^{2}

$- x^{2}$ with respect to

x

$x$ is

- \frac{d}{d x} [x^{2}]

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

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

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

Step 1.2.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 2

$n = 2$ .

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

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

Step 1.2.3

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

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

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

Step 1.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

4 x^{3} - 2 x + 1

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

4 x^{3} - 2 x + 1

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

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of

4 x^{3} - 2 x + 1

$4 x^{3} - 2 x + 1$ with respect to

x

$x$ is

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

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

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

Step 2.2

Evaluate

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

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Step 2.2.1

Since

$4$ is constant with respect to

$x$ , the derivative of

$4 x^{3}$ with respect to

$x$ is

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

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

Step 2.2.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 3$ .

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

Step 2.2.3

Multiply

$3$ by

$4$ .

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

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

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

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

Step 2.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.3.3

Multiply

$- 2$ by

$1$ .

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

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

$0$ .

$f'' (x) = 12 x^{2} - 2 + 0$

Step 2.4.2

Add

$12 x^{2} - 2$ and

$0$ .

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

$4 x^{3} - 2 x + 1 = 0$

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

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of

$x^{4} - x^{2} + x$ with respect to

$x$ is

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

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

Step 4.1.1.2

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 4$ .

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

Step 4.1.2

Evaluate

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

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Step 4.1.2.1

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- x^{2}$ with respect to

$x$ is

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

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

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

$n = 2$ .

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

Step 4.1.2.3

Multiply

$2$ by

$- 1$ .

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

Step 4.1.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 4.2

The first derivative of

$f (x)$ with respect to

$x$ is

$4 x^{3} - 2 x + 1$ .

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

Step 5

Set the first derivative equal to

$0$ then solve the equation

$4 x^{3} - 2 x + 1 = 0$ .

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Step 5.1

Set the first derivative equal to

$0$ .

$4 x^{3} - 2 x + 1 = 0$

Step 5.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx - 0.88464617$

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.

$x = - 0.88464617$

Step 8

Evaluate the second derivative at

$x = - 0.88464617$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(- 0.88464617)}^{2} - 2$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Raise

$- 0.88464617$ to the power of

$2$ .

$12 \cdot 0.78259885 - 2$

Step 9.1.2

Multiply

$12$ by

$0.78259885$ .

$9.3911863 - 2$

Step 9.2

Subtract

$2$ from

$9.3911863$ .

$7.3911863$

Step 10

$x = - 0.88464617$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 0.88464617$ is a local minimum

Step 11

Find the y-value when

$x = - 0.88464617$ .

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Step 11.1

Replace the variable

$x$ with

$- 0.88464617$ in the expression.

$f (- 0.88464617) = {(- 0.88464617)}^{4} - {(- 0.88464617)}^{2} - 0.88464617$

Step 11.2

Simplify the result.

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Step 11.2.1

Remove parentheses.

$f (- 0.88464617) = {(- 0.88464617)}^{4} - {(- 0.88464617)}^{2} - 0.88464617$

Step 11.2.2

Simplify each term.

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Step 11.2.2.1

Raise

$- 0.88464617$ to the power of

$4$ .

$f (- 0.88464617) = 0.61246097 - {(- 0.88464617)}^{2} - 0.88464617$

Step 11.2.2.2

Raise

$- 0.88464617$ to the power of

$2$ .

$f (- 0.88464617) = 0.61246097 - 1 \cdot 0.78259885 - 0.88464617$

Step 11.2.2.3

Multiply

$- 1$ by

$0.78259885$ .

$f (- 0.88464617) = 0.61246097 - 0.78259885 - 0.88464617$

Step 11.2.3

Simplify by subtracting numbers.

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Step 11.2.3.1

Subtract

$0.78259885$ from

$0.61246097$ .

$f (- 0.88464617) = - 0.17013788 - 0.88464617$

Step 11.2.3.2

Subtract

$0.88464617$ from

$- 0.17013788$ .

$f (- 0.88464617) = - 1.05478406$

Step 11.2.4

The final answer is

$- 1.05478406$ .

$y = - 1.05478406$

Step 12

These are the local extrema for

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

$(- 0.88464617, - 1.05478406)$ is a local minima

Step 13