Calculus Examples

$\frac{4 t}{3 t^{2} + 27}$

Step 1

Write $\frac{4 t}{3 t^{2} + 27}$ as a function.

$f (t) = \frac{4 t}{3 t^{2} + 27}$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Since $4$ is constant with respect to $t$ , the derivative of $\frac{4 t}{3 t^{2} + 27}$ with respect to $t$ is $4 \frac{d}{d t} [\frac{t}{3 t^{2} + 27}]$ .

$4 \frac{d}{d t} [\frac{t}{3 t^{2} + 27}]$

Step 2.1.2

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t$ and $g (t) = 3 t^{2} + 27$ .

$4 \frac{(3 t^{2} + 27) \frac{d}{d t} [t] - t \frac{d}{d t} [3 t^{2} + 27]}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3

Differentiate.

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Step 2.1.3.1

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$4 \frac{(3 t^{2} + 27) \cdot 1 - t \frac{d}{d t} [3 t^{2} + 27]}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.2

Multiply $3 t^{2} + 27$ by $1$ .

$4 \frac{3 t^{2} + 27 - t \frac{d}{d t} [3 t^{2} + 27]}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.3

By the Sum Rule, the derivative of $3 t^{2} + 27$ with respect to $t$ is $\frac{d}{d t} [3 t^{2}] + \frac{d}{d t} [27]$ .

$4 \frac{3 t^{2} + 27 - t (\frac{d}{d t} [3 t^{2}] + \frac{d}{d t} [27])}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.4

Since $3$ is constant with respect to $t$ , the derivative of $3 t^{2}$ with respect to $t$ is $3 \frac{d}{d t} [t^{2}]$ .

$4 \frac{3 t^{2} + 27 - t (3 \frac{d}{d t} [t^{2}] + \frac{d}{d t} [27])}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.5

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$4 \frac{3 t^{2} + 27 - t (3 (2 t) + \frac{d}{d t} [27])}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.6

Multiply $2$ by $3$ .

$4 \frac{3 t^{2} + 27 - t (6 t + \frac{d}{d t} [27])}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.7

Since $27$ is constant with respect to $t$ , the derivative of $27$ with respect to $t$ is $0$ .

$4 \frac{3 t^{2} + 27 - t (6 t + 0)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.8

Simplify the expression.

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Step 2.1.3.8.1

Add $6 t$ and $0$ .

$4 \frac{3 t^{2} + 27 - t (6 t)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.3.8.2

Multiply $6$ by $- 1$ .

$4 \frac{3 t^{2} + 27 - 6 t \cdot t}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.4

Raise $t$ to the power of $1$ .

$4 \frac{3 t^{2} + 27 - 6 (t^{1} t)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.5

Raise $t$ to the power of $1$ .

$4 \frac{3 t^{2} + 27 - 6 (t^{1} t^{1})}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 \frac{3 t^{2} + 27 - 6 t^{1 + 1}}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.7

Add $1$ and $1$ .

$4 \frac{3 t^{2} + 27 - 6 t^{2}}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.8

Subtract $6 t^{2}$ from $3 t^{2}$ .

$4 \frac{- 3 t^{2} + 27}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.9

Combine $4$ and $\frac{- 3 t^{2} + 27}{{(3 t^{2} + 27)}^{2}}$ .

$\frac{4 (- 3 t^{2} + 27)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10

Simplify.

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Step 2.1.10.1

Apply the distributive property.

$\frac{4 (- 3 t^{2}) + 4 \cdot 27}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.2

Simplify each term.

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Step 2.1.10.2.1

Multiply $- 3$ by $4$ .

$\frac{- 12 t^{2} + 4 \cdot 27}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.2.2

Multiply $4$ by $27$ .

$\frac{- 12 t^{2} + 108}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3

Simplify the numerator.

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Step 2.1.10.3.1

Factor $12$ out of $- 12 t^{2} + 108$ .

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Step 2.1.10.3.1.1

Factor $12$ out of $- 12 t^{2}$ .

$\frac{12 (- t^{2}) + 108}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3.1.2

Factor $12$ out of $108$ .

$\frac{12 (- t^{2}) + 12 (9)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3.1.3

Factor $12$ out of $12 (- t^{2}) + 12 (9)$ .

$\frac{12 (- t^{2} + 9)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3.2

Rewrite $9$ as $3^{2}$ .

$\frac{12 (- t^{2} + 3^{2})}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3.3

Reorder $- t^{2}$ and $3^{2}$ .

$\frac{12 (3^{2} - t^{2})}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.3.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3$ and $b = t$ .

$\frac{12 (3 + t) (3 - t)}{{(3 t^{2} + 27)}^{2}}$

Step 2.1.10.4

Simplify the denominator.

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Step 2.1.10.4.1

Factor $3$ out of $3 t^{2} + 27$ .

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Step 2.1.10.4.1.1

Factor $3$ out of $3 t^{2}$ .

$\frac{12 (3 + t) (3 - t)}{{(3 (t^{2}) + 27)}^{2}}$

Step 2.1.10.4.1.2

Factor $3$ out of $27$ .

$\frac{12 (3 + t) (3 - t)}{{(3 t^{2} + 3 \cdot 9)}^{2}}$

Step 2.1.10.4.1.3

Factor $3$ out of $3 t^{2} + 3 \cdot 9$ .

$\frac{12 (3 + t) (3 - t)}{{(3 (t^{2} + 9))}^{2}}$

Step 2.1.10.4.2

Apply the product rule to $3 (t^{2} + 9)$ .

$\frac{12 (3 + t) (3 - t)}{3^{2} {(t^{2} + 9)}^{2}}$

Step 2.1.10.4.3

Raise $3$ to the power of $2$ .

$\frac{12 (3 + t) (3 - t)}{9 {(t^{2} + 9)}^{2}}$

Step 2.1.10.5

Cancel the common factor of $12$ and $9$ .

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Step 2.1.10.5.1

Factor $3$ out of $12 (3 + t) (3 - t)$ .

$\frac{3 (4 (3 + t) (3 - t))}{9 {(t^{2} + 9)}^{2}}$

Step 2.1.10.5.2

Cancel the common factors.

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Step 2.1.10.5.2.1

Factor $3$ out of $9 {(t^{2} + 9)}^{2}$ .

$\frac{3 (4 (3 + t) (3 - t))}{3 (3 {(t^{2} + 9)}^{2})}$

Step 2.1.10.5.2.2

Cancel the common factor.

$\frac{3 (4 (3 + t) (3 - t))}{3 (3 {(t^{2} + 9)}^{2})}$

Step 2.1.10.5.2.3

Rewrite the expression.

$f' (t) = \frac{4 (3 + t) (3 - t)}{3 {(t^{2} + 9)}^{2}}$

Step 2.2

The first derivative of $f (t)$ with respect to $t$ is $\frac{4 (3 + t) (3 - t)}{3 {(t^{2} + 9)}^{2}}$ .

$\frac{4 (3 + t) (3 - t)}{3 {(t^{2} + 9)}^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{4 (3 + t) (3 - t)}{3 {(t^{2} + 9)}^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{4 (3 + t) (3 - t)}{3 {(t^{2} + 9)}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$4 (3 + t) (3 - t) = 0$

Step 3.3

Solve the equation for $t$ .

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Step 3.3.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 + t = 0$

$3 - t = 0$

Step 3.3.2

Set $3 + t$ equal to $0$ and solve for $t$ .

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Step 3.3.2.1

Set $3 + t$ equal to $0$ .

$3 + t = 0$

Step 3.3.2.2

Subtract $3$ from both sides of the equation.

$t = - 3$

Step 3.3.3

Set $3 - t$ equal to $0$ and solve for $t$ .

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Step 3.3.3.1

Set $3 - t$ equal to $0$ .

$3 - t = 0$

Step 3.3.3.2

Solve $3 - t = 0$ for $t$ .

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Step 3.3.3.2.1

Subtract $3$ from both sides of the equation.

$- t = - 3$

Step 3.3.3.2.2

Divide each term in $- t = - 3$ by $- 1$ and simplify.

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Step 3.3.3.2.2.1

Divide each term in $- t = - 3$ by $- 1$ .

$\frac{- t}{- 1} = \frac{- 3}{- 1}$

Step 3.3.3.2.2.2

Simplify the left side.

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Step 3.3.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{t}{1} = \frac{- 3}{- 1}$

Step 3.3.3.2.2.2.2

Divide $t$ by $1$ .

$t = \frac{- 3}{- 1}$

Step 3.3.3.2.2.3

Simplify the right side.

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Step 3.3.3.2.2.3.1

Divide $- 3$ by $- 1$ .

$t = 3$

Step 3.3.4

The final solution is all the values that make $4 (3 + t) (3 - t) = 0$ true.

$t = - 3, 3$

Step 4

The values which make the derivative equal to $0$ are $- 3, 3$ .

$- 3, 3$

Step 5

Split $(- \infty, \infty)$ into separate intervals around the $t$ values that make the derivative $0$ or undefined.

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $t$ with $- 4$ in the expression.

$f' (- 4) = \frac{4 (3 - 4) (3 - (- 4))}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Remove parentheses.

$f' (- 4) = \frac{4 (3 - 4) (3 - (- 4))}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.2

Simplify the numerator.

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Step 6.2.2.1

Multiply $- 1$ by $- 4$ .

$f' (- 4) = \frac{4 (3 - 4) (3 + 4)}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.2.2

Subtract $4$ from $3$ .

$f' (- 4) = \frac{4 \cdot (- 1 (3 + 4))}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.2.3

Combine exponents.

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Step 6.2.2.3.1

Factor out negative.

$f' (- 4) = \frac{- (4 (3 + 4))}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.2.3.2

Multiply $4$ by $- 1$ .

$f' (- 4) = \frac{- 4 (3 + 4)}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.2.4

Add $3$ and $4$ .

$f' (- 4) = \frac{- 4 \cdot 7}{3 {({(- 4)}^{2} + 9)}^{2}}$

Step 6.2.3

Simplify the denominator.

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Step 6.2.3.1

Raise $- 4$ to the power of $2$ .

$f' (- 4) = \frac{- 4 \cdot 7}{3 {(16 + 9)}^{2}}$

Step 6.2.3.2

Add $16$ and $9$ .

$f' (- 4) = \frac{- 4 \cdot 7}{3 \cdot 25^{2}}$

Step 6.2.3.3

Raise $25$ to the power of $2$ .

$f' (- 4) = \frac{- 4 \cdot 7}{3 \cdot 625}$

Step 6.2.4

Simplify the expression.

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Step 6.2.4.1

Multiply $- 4$ by $7$ .

$f' (- 4) = \frac{- 28}{3 \cdot 625}$

Step 6.2.4.2

Multiply $3$ by $625$ .

$f' (- 4) = \frac{- 28}{1875}$

Step 6.2.4.3

Move the negative in front of the fraction.

$f' (- 4) = - \frac{28}{1875}$

Step 6.2.5

The final answer is $- \frac{28}{1875}$ .

$- \frac{28}{1875}$

Step 6.3

At $t = - 4$ the derivative is $- \frac{28}{1875}$ . Since this is negative, the function is decreasing on $(- \infty, - 3)$ .

Decreasing on $(- \infty, - 3)$ since $f' (t) < 0$

Step 7

Substitute a value from the interval $(- 3, 3)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $t$ with $0$ in the expression.

$f' (0) = \frac{4 (3 + 0) (3 - (0))}{3 {({(0)}^{2} + 9)}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Reduce the expression by cancelling the common factors.

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Step 7.2.1.1

Remove parentheses.

$f' (0) = \frac{4 (3 + 0) (3 - (0))}{3 {({(0)}^{2} + 9)}^{2}}$

Step 7.2.1.2

Cancel the common factor of $3 + 0$ and $3$ .

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Step 7.2.1.2.1

Factor $3$ out of $4 (3 + 0) (3 - (0))$ .

$f' (0) = \frac{3 (4 (1 + 0) (3 - (0)))}{3 {({(0)}^{2} + 9)}^{2}}$

Step 7.2.1.2.2

Cancel the common factors.

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Step 7.2.1.2.2.1

Factor $3$ out of $3 {({(0)}^{2} + 9)}^{2}$ .

$f' (0) = \frac{3 (4 (1 + 0) (3 - (0)))}{3 ({({(0)}^{2} + 9)}^{2})}$

Step 7.2.1.2.2.2

Cancel the common factor.

$f' (0) = \frac{3 (4 (1 + 0) (3 - (0)))}{3 {({(0)}^{2} + 9)}^{2}}$

Step 7.2.1.2.2.3

Rewrite the expression.

$f' (0) = \frac{4 (1 + 0) (3 - (0))}{{({(0)}^{2} + 9)}^{2}}$

Step 7.2.2

Simplify the numerator.

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Step 7.2.2.1

Multiply $- 1$ by $0$ .

$f' (0) = \frac{4 (1 + 0) (3 + 0)}{{(0^{2} + 9)}^{2}}$

Step 7.2.2.2

Add $1$ and $0$ .

$f' (0) = \frac{4 \cdot (1 (3 + 0))}{{(0^{2} + 9)}^{2}}$

Step 7.2.2.3

Multiply $4$ by $1$ .

$f' (0) = \frac{4 (3 + 0)}{{(0^{2} + 9)}^{2}}$

Step 7.2.2.4

Add $3$ and $0$ .

$f' (0) = \frac{4 \cdot 3}{{(0^{2} + 9)}^{2}}$

Step 7.2.3

Simplify the denominator.

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Step 7.2.3.1

Raising $0$ to any positive power yields $0$ .

$f' (0) = \frac{4 \cdot 3}{{(0 + 9)}^{2}}$

Step 7.2.3.2

Add $0$ and $9$ .

$f' (0) = \frac{4 \cdot 3}{9^{2}}$

Step 7.2.3.3

Raise $9$ to the power of $2$ .

$f' (0) = \frac{4 \cdot 3}{81}$

Step 7.2.4

Reduce the expression by cancelling the common factors.

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Step 7.2.4.1

Multiply $4$ by $3$ .

$f' (0) = \frac{12}{81}$

Step 7.2.4.2

Cancel the common factor of $12$ and $81$ .

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Step 7.2.4.2.1

Factor $3$ out of $12$ .

$f' (0) = \frac{3 (4)}{81}$

Step 7.2.4.2.2

Cancel the common factors.

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Step 7.2.4.2.2.1

Factor $3$ out of $81$ .

$f' (0) = \frac{3 \cdot 4}{3 \cdot 27}$

Step 7.2.4.2.2.2

Cancel the common factor.

$f' (0) = \frac{3 \cdot 4}{3 \cdot 27}$

Step 7.2.4.2.2.3

Rewrite the expression.

$f' (0) = \frac{4}{27}$

Step 7.2.5

The final answer is $\frac{4}{27}$ .

$\frac{4}{27}$

Step 7.3

At $t = 0$ the derivative is $\frac{4}{27}$ . Since this is positive, the function is increasing on $(- 3, 3)$ .

Increasing on $(- 3, 3)$ since $f' (t) > 0$

Step 8

Substitute a value from the interval $(3, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $t$ with $4$ in the expression.

$f' (4) = \frac{4 (3 + 4) (3 - (4))}{3 {({(4)}^{2} + 9)}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Remove parentheses.

$f' (4) = \frac{4 (3 + 4) (3 - (4))}{3 {({(4)}^{2} + 9)}^{2}}$

Step 8.2.2

Simplify the numerator.

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Step 8.2.2.1

Multiply $- 1$ by $4$ .

$f' (4) = \frac{4 (3 + 4) (3 - 4)}{3 {(4^{2} + 9)}^{2}}$

Step 8.2.2.2

Add $3$ and $4$ .

$f' (4) = \frac{4 \cdot (7 (3 - 4))}{3 {(4^{2} + 9)}^{2}}$

Step 8.2.2.3

Multiply $4$ by $7$ .

$f' (4) = \frac{28 (3 - 4)}{3 {(4^{2} + 9)}^{2}}$

Step 8.2.2.4

Subtract $4$ from $3$ .

$f' (4) = \frac{28 \cdot - 1}{3 {(4^{2} + 9)}^{2}}$

Step 8.2.3

Simplify the denominator.

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Step 8.2.3.1

Raise $4$ to the power of $2$ .

$f' (4) = \frac{28 \cdot - 1}{3 {(16 + 9)}^{2}}$

Step 8.2.3.2

Add $16$ and $9$ .

$f' (4) = \frac{28 \cdot - 1}{3 \cdot 25^{2}}$

Step 8.2.3.3

Raise $25$ to the power of $2$ .

$f' (4) = \frac{28 \cdot - 1}{3 \cdot 625}$

Step 8.2.4

Simplify the expression.

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Step 8.2.4.1

Multiply $28$ by $- 1$ .

$f' (4) = \frac{- 28}{3 \cdot 625}$

Step 8.2.4.2

Multiply $3$ by $625$ .

$f' (4) = \frac{- 28}{1875}$

Step 8.2.4.3

Move the negative in front of the fraction.

$f' (4) = - \frac{28}{1875}$

Step 8.2.5

The final answer is $- \frac{28}{1875}$ .

$- \frac{28}{1875}$

Step 8.3

At $t = 4$ the derivative is $- \frac{28}{1875}$ . Since this is negative, the function is decreasing on $(3, \infty)$ .

Decreasing on $(3, \infty)$ since $f' (t) < 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- 3, 3)$

Decreasing on: $(- \infty, - 3), (3, \infty)$

Step 10