Calculus Examples

f (x) = 8 - x

$f (x) = 8 - x$ ,

(- 3, 5)

$(- 3, 5)$

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

Differentiate.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of

8 - x

$8 - x$ with respect to

x

$x$ is

\frac{d}{d x} [8] + \frac{d}{d x} [- x]

$\frac{d}{d x} [8] + \frac{d}{d x} [- x]$ .

\frac{d}{d x} [8] + \frac{d}{d x} [- x]

$\frac{d}{d x} [8] + \frac{d}{d x} [- x]$

Step 1.1.1.1.2

Since

8

$8$ is constant with respect to

x

$x$ , the derivative of

8

$8$ with respect to

x

$x$ is

0

$0$ .

0 + \frac{d}{d x} [- x]

$0 + \frac{d}{d x} [- x]$

0 + \frac{d}{d x} [- x]

$0 + \frac{d}{d x} [- x]$

Step 1.1.1.2

Evaluate

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

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

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Step 1.1.1.2.1

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- x

$- x$ with respect to

x

$x$ is

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

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

0 - \frac{d}{d x} [x]

$0 - \frac{d}{d x} [x]$

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 = 1

$n = 1$ .

0 - 1 \cdot 1

$0 - 1 \cdot 1$

Step 1.1.1.2.3

Multiply

- 1

$- 1$ by

1

$1$ .

0 - 1

$0 - 1$

0 - 1

$0 - 1$

Step 1.1.1.3

Subtract

1

$1$ from

0

$0$ .

$f' (x) = - 1$

Step 1.1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$- 1$ .

$- 1$

Step 1.2

Set the first derivative equal to

$0$ then solve the equation

$- 1 = 0$ .

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Step 1.2.1

Set the first derivative equal to

$0$ .

$- 1 = 0$

Step 1.2.2

Since

$- 1 \neq 0$ , there are no solutions.

No solution

Step 1.3

There are no values of

$x$ in the domain of the original problem where the derivative is

$0$ or undefined.

No critical points found

Step 2

Since there is no value of

$x$ that makes the first derivative equal to

$0$ , there are no local extrema.

No Local Extrema

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.

No absolute maximum

No absolute minimum

Step 4