Calculus Examples

f (x) = - x^{2} + 10

$f (x) = - x^{2} + 10$ on

- 3

$- 3$ ,

4

$4$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

By the Sum Rule, the derivative of

- x^{2} + 10

$- x^{2} + 10$ with respect to

x

$x$ is

\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [10]

$\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [10]$ .

\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [10]

$\frac{d}{d x} [- x^{2}] + \frac{d}{d x} [10]$

Step 1.1.1.2

Evaluate

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

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

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Step 1.1.1.2.1

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}]

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

- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [10]

$- \frac{d}{d x} [x^{2}] + \frac{d}{d x} [10]$

Step 1.1.1.2.2

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

$n = 2$ .

- (2 x) + \frac{d}{d x} [10]

$- (2 x) + \frac{d}{d x} [10]$

Step 1.1.1.2.3

Multiply

2

$2$ by

- 1

$- 1$ .

- 2 x + \frac{d}{d x} [10]

$- 2 x + \frac{d}{d x} [10]$

- 2 x + \frac{d}{d x} [10]

$- 2 x + \frac{d}{d x} [10]$

Step 1.1.1.3

Differentiate using the Constant Rule.

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Step 1.1.1.3.1

Since

10

$10$ is constant with respect to

x

$x$ , the derivative of

10

$10$ with respect to

x

$x$ is

0

$0$ .

- 2 x + 0

$- 2 x + 0$

Step 1.1.1.3.2

Add

- 2 x

$- 2 x$ and

0

$0$ .

f' (x) = - 2 x

$f' (x) = - 2 x$

f' (x) = - 2 x

$f' (x) = - 2 x$

f' (x) = - 2 x

$f' (x) = - 2 x$

Step 1.1.2

The first derivative of

f (x)

$f (x)$ with respect to

x

$x$ is

- 2 x

$- 2 x$ .

- 2 x

$- 2 x$

- 2 x

$- 2 x$

Step 1.2

Set the first derivative equal to

0

$0$ then solve the equation

- 2 x = 0

$- 2 x = 0$ .

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Step 1.2.1

Set the first derivative equal to

0

$0$ .

- 2 x = 0

$- 2 x = 0$

Step 1.2.2

Divide each term in

- 2 x = 0

$- 2 x = 0$ by

- 2

$- 2$ and simplify.

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Step 1.2.2.1

Divide each term in

- 2 x = 0

$- 2 x = 0$ by

- 2

$- 2$ .

\frac{- 2 x}{- 2} = \frac{0}{- 2}

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of

- 2

$- 2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Step 1.2.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{0}{- 2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide

$0$ by

$- 2$ .

$x = 0$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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 1.4

Evaluate

$- x^{2} + 10$ at each

$x$ value where the derivative is

$0$ or undefined.

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Step 1.4.1

Evaluate at

$x = 0$ .

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Step 1.4.1.1

Substitute

$0$ for

$x$ .

$- {(0)}^{2} + 10$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

Raising

$0$ to any positive power yields

$0$ .

$- 0 + 10$

Step 1.4.1.2.1.2

Multiply

$- 1$ by

$0$ .

$0 + 10$

Step 1.4.1.2.2

Add

$0$ and

$10$ .

$10$

Step 1.4.2

List all of the points.

$(0, 10)$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at

$x = - 3$ .

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Step 2.1.1

Substitute

$- 3$ for

$x$ .

$- {(- 3)}^{2} + 10$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Raise

$- 3$ to the power of

$2$ .

$- 1 \cdot 9 + 10$

Step 2.1.2.1.2

Multiply

$- 1$ by

$9$ .

$- 9 + 10$

Step 2.1.2.2

Add

$- 9$ and

$10$ .

$1$

Step 2.2

Evaluate at

$x = 4$ .

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Step 2.2.1

Substitute

$4$ for

$x$ .

$- {(4)}^{2} + 10$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Raise

$4$ to the power of

$2$ .

$- 1 \cdot 16 + 10$

Step 2.2.2.1.2

Multiply

$- 1$ by

$16$ .

$- 16 + 10$

Step 2.2.2.2

Add

$- 16$ and

$10$ .

$- 6$

Step 2.3

List all of the points.

$(- 3, 1), (4, - 6)$

Step 3

Compare the

$f (x)$ values found for each value of

$x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest

$f (x)$ value and the minimum will occur at the lowest

$f (x)$ value.

Absolute Maximum:

$(0, 10)$

Absolute Minimum:

$(4, - 6)$

Step 4