Calculus Examples

g (t) = \frac{t}{t - 8}

$g (t) = \frac{t}{t - 8}$ on

10

$10$ ,

12

$12$

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 using the Quotient Rule which states that

\frac{d}{d t} [\frac{f (t)}{g (t)}]

$\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is

\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}

$\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where

f (t) = t

$f (t) = t$ and

g (t) = t - 8

$g (t) = t - 8$ .

\frac{(t - 8) \frac{d}{d t} [t] - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}

$\frac{(t - 8) \frac{d}{d t} [t] - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

Differentiate using the Power Rule which states that

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

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

n t^{n - 1}

$n t^{n - 1}$ where

n = 1

$n = 1$ .

\frac{(t - 8) \cdot 1 - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}

$\frac{(t - 8) \cdot 1 - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}$

Step 1.1.1.2.2

Multiply

t - 8

$t - 8$ by

1

$1$ .

\frac{t - 8 - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}

$\frac{t - 8 - t \frac{d}{d t} [t - 8]}{{(t - 8)}^{2}}$

Step 1.1.1.2.3

By the Sum Rule, the derivative of

t - 8

$t - 8$ with respect to

t

$t$ is

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

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

\frac{t - 8 - t (\frac{d}{d t} [t] + \frac{d}{d t} [- 8])}{{(t - 8)}^{2}}

$\frac{t - 8 - t (\frac{d}{d t} [t] + \frac{d}{d t} [- 8])}{{(t - 8)}^{2}}$

Step 1.1.1.2.4

Differentiate using the Power Rule which states that

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

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

n t^{n - 1}

$n t^{n - 1}$ where

n = 1

$n = 1$ .

\frac{t - 8 - t (1 + \frac{d}{d t} [- 8])}{{(t - 8)}^{2}}

$\frac{t - 8 - t (1 + \frac{d}{d t} [- 8])}{{(t - 8)}^{2}}$

Step 1.1.1.2.5

Since

- 8

$- 8$ is constant with respect to

t

$t$ , the derivative of

- 8

$- 8$ with respect to

t

$t$ is

0

$0$ .

\frac{t - 8 - t (1 + 0)}{{(t - 8)}^{2}}

$\frac{t - 8 - t (1 + 0)}{{(t - 8)}^{2}}$

Step 1.1.1.2.6

Simplify by adding terms.

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Step 1.1.1.2.6.1

Add

1

$1$ and

0

$0$ .

\frac{t - 8 - t \cdot 1}{{(t - 8)}^{2}}

$\frac{t - 8 - t \cdot 1}{{(t - 8)}^{2}}$

Step 1.1.1.2.6.2

Multiply

- 1

$- 1$ by

1

$1$ .

\frac{t - 8 - t}{{(t - 8)}^{2}}

$\frac{t - 8 - t}{{(t - 8)}^{2}}$

Step 1.1.1.2.6.3

Subtract

t

$t$ from

t

$t$ .

\frac{0 - 8}{{(t - 8)}^{2}}

$\frac{0 - 8}{{(t - 8)}^{2}}$

Step 1.1.1.2.6.4

Simplify the expression.

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Step 1.1.1.2.6.4.1

Subtract

8

$8$ from

0

$0$ .

\frac{- 8}{{(t - 8)}^{2}}

$\frac{- 8}{{(t - 8)}^{2}}$

Step 1.1.1.2.6.4.2

Move the negative in front of the fraction.

f' (t) = - \frac{8}{{(t - 8)}^{2}}

$f' (t) = - \frac{8}{{(t - 8)}^{2}}$

f' (t) = - \frac{8}{{(t - 8)}^{2}}

$f' (t) = - \frac{8}{{(t - 8)}^{2}}$

f' (t) = - \frac{8}{{(t - 8)}^{2}}

$f' (t) = - \frac{8}{{(t - 8)}^{2}}$

f' (t) = - \frac{8}{{(t - 8)}^{2}}

$f' (t) = - \frac{8}{{(t - 8)}^{2}}$

f' (t) = - \frac{8}{{(t - 8)}^{2}}

$f' (t) = - \frac{8}{{(t - 8)}^{2}}$

Step 1.1.2

The first derivative of

g (t)

$g (t)$ with respect to

t

$t$ is

- \frac{8}{{(t - 8)}^{2}}

$- \frac{8}{{(t - 8)}^{2}}$ .

- \frac{8}{{(t - 8)}^{2}}

$- \frac{8}{{(t - 8)}^{2}}$

- \frac{8}{{(t - 8)}^{2}}

$- \frac{8}{{(t - 8)}^{2}}$

Step 1.2

Set the first derivative equal to

0

$0$ then solve the equation

- \frac{8}{{(t - 8)}^{2}} = 0

$- \frac{8}{{(t - 8)}^{2}} = 0$ .

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Step 1.2.1

Set the first derivative equal to

0

$0$ .

- \frac{8}{{(t - 8)}^{2}} = 0

$- \frac{8}{{(t - 8)}^{2}} = 0$

Step 1.2.2

Set the numerator equal to zero.

8 = 0

$8 = 0$

Step 1.2.3

Since

8 \neq 0

$8 \neq 0$ , there are no solutions.

No solution

No solution

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the denominator in

\frac{8}{{(t - 8)}^{2}}

$\frac{8}{{(t - 8)}^{2}}$ equal to

0

$0$ to find where the expression is undefined.

{(t - 8)}^{2} = 0

${(t - 8)}^{2} = 0$

Step 1.3.2

Solve for

t

$t$ .

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Step 1.3.2.1

Set the

t - 8

$t - 8$ equal to

0

$0$ .

t - 8 = 0

$t - 8 = 0$

Step 1.3.2.2

Add

8

$8$ to both sides of the equation.

t = 8

$t = 8$

t = 8

$t = 8$

t = 8

$t = 8$

Step 1.4

Evaluate

\frac{t}{t - 8}

$\frac{t}{t - 8}$ at each

t

$t$ value where the derivative is

0

$0$ or undefined.

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Step 1.4.1

Evaluate at

t = 8

$t = 8$ .

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Step 1.4.1.1

Substitute

8

$8$ for

t

$t$ .

\frac{8}{(8) - 8}

$\frac{8}{(8) - 8}$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Subtract

8

$8$ from

8

$8$ .

\frac{8}{0}

$\frac{8}{0}$

Step 1.4.1.2.2

The expression contains a division by

0

$0$ . The expression is undefined.

Undefined

Undefined

Undefined

Undefined

Step 1.5

There are no values of

t

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

0

$0$ or undefined.

No critical points found

No critical points found

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at

t = 10

$t = 10$ .

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Step 2.1.1

Substitute

10

$10$ for

t

$t$ .

\frac{10}{(10) - 8}

$\frac{10}{(10) - 8}$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Cancel the common factor of

10

$10$ and

(10) - 8

$(10) - 8$ .

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Step 2.1.2.1.1

Factor

2

$2$ out of

10

$10$ .

\frac{2 (5)}{(10) - 8}

$\frac{2 (5)}{(10) - 8}$

Step 2.1.2.1.2

Cancel the common factors.

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Step 2.1.2.1.2.1

Factor

2

$2$ out of

10

$10$ .

\frac{2 \cdot 5}{2 \cdot 5 - 8}

$\frac{2 \cdot 5}{2 \cdot 5 - 8}$

Step 2.1.2.1.2.2

Factor

2

$2$ out of

- 8

$- 8$ .

\frac{2 (5)}{2 \cdot 5 + 2 \cdot - 4}

$\frac{2 (5)}{2 \cdot 5 + 2 \cdot - 4}$

Step 2.1.2.1.2.3

Factor

2

$2$ out of

2 \cdot 5 + 2 \cdot - 4

$2 \cdot 5 + 2 \cdot - 4$ .

\frac{2 (5)}{2 \cdot (5 - 4)}

$\frac{2 (5)}{2 \cdot (5 - 4)}$

Step 2.1.2.1.2.4

Cancel the common factor.

$\frac{2 \cdot 5}{2 \cdot (5 - 4)}$

Step 2.1.2.1.2.5

Rewrite the expression.

$\frac{5}{5 - 4}$

$\frac{5}{5 - 4}$

$\frac{5}{5 - 4}$

Step 2.1.2.2

Simplify the expression.

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Step 2.1.2.2.1

Subtract

$4$ from

$5$ .

$\frac{5}{1}$

Step 2.1.2.2.2

Divide

$5$ by

$1$ .

$5$

$5$

$5$

$5$

Step 2.2

Evaluate at

$t = 12$ .

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Step 2.2.1

Substitute

$12$ for

$t$ .

$\frac{12}{(12) - 8}$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Cancel the common factor of

$12$ and

$(12) - 8$ .

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Step 2.2.2.1.1

Factor

$4$ out of

$12$ .

$\frac{4 (3)}{(12) - 8}$

Step 2.2.2.1.2

Cancel the common factors.

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Step 2.2.2.1.2.1

Factor

$4$ out of

$12$ .

$\frac{4 \cdot 3}{4 \cdot 3 - 8}$

Step 2.2.2.1.2.2

Factor

$4$ out of

$- 8$ .

$\frac{4 (3)}{4 \cdot 3 + 4 \cdot - 2}$

Step 2.2.2.1.2.3

Factor

$4$ out of

$4 \cdot 3 + 4 \cdot - 2$ .

$\frac{4 (3)}{4 \cdot (3 - 2)}$

Step 2.2.2.1.2.4

Cancel the common factor.

$\frac{4 \cdot 3}{4 \cdot (3 - 2)}$

Step 2.2.2.1.2.5

Rewrite the expression.

$\frac{3}{3 - 2}$

$\frac{3}{3 - 2}$

$\frac{3}{3 - 2}$

Step 2.2.2.2

Simplify the expression.

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Step 2.2.2.2.1

Subtract

$2$ from

$3$ .

$\frac{3}{1}$

Step 2.2.2.2.2

Divide

$3$ by

$1$ .

$3$

$3$

$3$

$3$

Step 2.3

List all of the points.

$(10, 5), (12, 3)$

$(10, 5), (12, 3)$

Step 3

Compare the

$g (t)$ values found for each value of

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

$g (t)$ value and the minimum will occur at the lowest

$g (t)$ value.

Absolute Maximum:

$(10, 5)$

Absolute Minimum:

$(12, 3)$

Step 4

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$