Calculus Examples

f (x) = \frac{2 x + 5}{3}

$f (x) = \frac{2 x + 5}{3}$ ,

[0, 5]

$[0, 5]$

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

Since

\frac{1}{3}

$\frac{1}{3}$ is constant with respect to

x

$x$ , the derivative of

\frac{2 x + 5}{3}

$\frac{2 x + 5}{3}$ with respect to

x

$x$ is

\frac{1}{3} \frac{d}{d x} [2 x + 5]

$\frac{1}{3} \frac{d}{d x} [2 x + 5]$ .

\frac{1}{3} \frac{d}{d x} [2 x + 5]

$\frac{1}{3} \frac{d}{d x} [2 x + 5]$

Step 1.1.1.2

By the Sum Rule, the derivative of

2 x + 5

$2 x + 5$ with respect to

x

$x$ is

\frac{d}{d x} [2 x] + \frac{d}{d x} [5]

$\frac{d}{d x} [2 x] + \frac{d}{d x} [5]$ .

\frac{1}{3} (\frac{d}{d x} [2 x] + \frac{d}{d x} [5])

$\frac{1}{3} (\frac{d}{d x} [2 x] + \frac{d}{d x} [5])$

Step 1.1.1.3

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

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

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

\frac{1}{3} (2 \frac{d}{d x} [x] + \frac{d}{d x} [5])

$\frac{1}{3} (2 \frac{d}{d x} [x] + \frac{d}{d x} [5])$

Step 1.1.1.4

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

$n = 1$ .

\frac{1}{3} (2 \cdot 1 + \frac{d}{d x} [5])

$\frac{1}{3} (2 \cdot 1 + \frac{d}{d x} [5])$

Step 1.1.1.5

Multiply

2

$2$ by

1

$1$ .

\frac{1}{3} (2 + \frac{d}{d x} [5])

$\frac{1}{3} (2 + \frac{d}{d x} [5])$

Step 1.1.1.6

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5

$5$ with respect to

x

$x$ is

0

$0$ .

\frac{1}{3} (2 + 0)

$\frac{1}{3} (2 + 0)$

Step 1.1.1.7

Combine fractions.

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Step 1.1.1.7.1

Add

2

$2$ and

0

$0$ .

\frac{1}{3} \cdot 2

$\frac{1}{3} \cdot 2$

Step 1.1.1.7.2

Combine

\frac{1}{3}

$\frac{1}{3}$ and

2

$2$ .

$f' (x) = \frac{2}{3}$

Step 1.1.2

The first derivative of

$f (x)$ with respect to

$x$ is

$\frac{2}{3}$ .

$\frac{2}{3}$

Step 1.2

Set the first derivative equal to

$0$ then solve the equation

$\frac{2}{3} = 0$ .

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Step 1.2.1

Set the first derivative equal to

$0$ .

$\frac{2}{3} = 0$

Step 1.2.2

Set the numerator equal to zero.

$2 = 0$

Step 1.2.3

Since

$2 \neq 0$ , 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

Evaluate at the included endpoints.

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Step 2.1

Evaluate at

$x = 0$ .

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Step 2.1.1

Substitute

$0$ for

$x$ .

$\frac{2 (0) + 5}{3}$

Step 2.1.2

Simplify the numerator.

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Step 2.1.2.1

Multiply

$2$ by

$0$ .

$\frac{0 + 5}{3}$

Step 2.1.2.2

Add

$0$ and

$5$ .

$\frac{5}{3}$

Step 2.2

Evaluate at

$x = 5$ .

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Step 2.2.1

Substitute

$5$ for

$x$ .

$\frac{2 (5) + 5}{3}$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Simplify the numerator.

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Step 2.2.2.1.1

Multiply

$2$ by

$5$ .

$\frac{10 + 5}{3}$

Step 2.2.2.1.2

Add

$10$ and

$5$ .

$\frac{15}{3}$

Step 2.2.2.2

Divide

$15$ by

$3$ .

$5$

Step 2.3

List all of the points.

$(0, \frac{5}{3}), (5, 5)$

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:

$(5, 5)$

Absolute Minimum:

$(0, \frac{5}{3})$

Step 4