Calculus Examples

${(t^{2} - 4)}^{3}$

Step 1

Write ${(t^{2} - 4)}^{3}$ as a function.

$f (t) = {(t^{2} - 4)}^{3}$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{3}$ and $g (t) = t^{2} - 4$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $t^{2} - 4$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d t} [t^{2} - 4]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$3 u^{2} \frac{d}{d t} [t^{2} - 4]$

Step 2.1.3

Replace all occurrences of $u$ with $t^{2} - 4$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $t^{2} - 4$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 4]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Since $- 4$ is constant with respect to $t$ , the derivative of $- 4$ with respect to $t$ is $0$ .

$3 {(t^{2} - 4)}^{2} (2 t + 0)$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $2 t$ and $0$ .

$3 {(t^{2} - 4)}^{2} (2 t)$

Step 2.2.4.2

Multiply $2$ by $3$ .

$6 {(t^{2} - 4)}^{2} t$

Step 2.2.4.3

Reorder the factors of $6 {(t^{2} - 4)}^{2} t$ .

$6 t {(t^{2} - 4)}^{2}$

Step 3

Find the second derivative of the function.

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Step 3.1

Since $6$ is constant with respect to $t$ , the derivative of $6 t {(t^{2} - 4)}^{2}$ with respect to $t$ is $6 \frac{d}{d t} [t {(t^{2} - 4)}^{2}]$ .

$f'' (t) = 6 \frac{d}{d t} (t {(t^{2} - 4)}^{2})$

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = {(t^{2} - 4)}^{2}$ .

$f'' (t) = 6 (t \frac{d}{d t} ({(t^{2} - 4)}^{2}) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = t^{2} - 4$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $t^{2} - 4$ .

$f'' (t) = 6 (t (\frac{d}{d u} (u^{2}) \frac{d}{d t} (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.3.2

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

$f'' (t) = 6 (t (2 u \frac{d}{d t} (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.3.3

Replace all occurrences of $u$ with $t^{2} - 4$ .

$f'' (t) = 6 (t (2 (t^{2} - 4) \frac{d}{d t} (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.4

Differentiate.

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Step 3.4.1

By the Sum Rule, the derivative of $t^{2} - 4$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 4]$ .

$f'' (t) = 6 (t (2 (t^{2} - 4) (\frac{d}{d t} (t^{2}) + \frac{d}{d t} (- 4))) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.4.2

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

$f'' (t) = 6 (t (2 (t^{2} - 4) (2 t + \frac{d}{d t} (- 4))) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.4.3

Since $- 4$ is constant with respect to $t$ , the derivative of $- 4$ with respect to $t$ is $0$ .

$f'' (t) = 6 (t (2 (t^{2} - 4) (2 t + 0)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.4.4

Simplify the expression.

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Step 3.4.4.1

Add $2 t$ and $0$ .

$f'' (t) = 6 (t (2 (t^{2} - 4) (2 t)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.4.4.2

Multiply $2$ by $2$ .

$f'' (t) = 6 (t (4 (t^{2} - 4) t) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.5

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

$f'' (t) = 6 (t \cdot t (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.6

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

$f'' (t) = 6 (t \cdot t (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.7

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

$f'' (t) = 6 (t^{1 + 1} (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.8

Add $1$ and $1$ .

$f'' (t) = 6 (t^{2} (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2} \frac{d}{d t} (t))$

Step 3.9

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

$f'' (t) = 6 (t^{2} (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2} \cdot 1)$

Step 3.10

Multiply ${(t^{2} - 4)}^{2}$ by $1$ .

$f'' (t) = 6 (t^{2} (4 (t^{2} - 4)) + {(t^{2} - 4)}^{2})$

Step 3.11

Simplify.

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Step 3.11.1

Apply the distributive property.

$f'' (t) = 6 (t^{2} (4 t^{2} + 4 \cdot - 4) + {(t^{2} - 4)}^{2})$

Step 3.11.2

Apply the distributive property.

$f'' (t) = 6 (t^{2} (4 t^{2}) + t^{2} (4 \cdot - 4) + {(t^{2} - 4)}^{2})$

Step 3.11.3

Apply the distributive property.

$f'' (t) = 6 (t^{2} (4 t^{2})) + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4

Combine terms.

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Step 3.11.4.1

Multiply $t^{2}$ by $t^{2}$ by adding the exponents.

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Step 3.11.4.1.1

Move $t^{2}$ .

$f'' (t) = 6 (t^{2} t^{2} \cdot 4) + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.1.2

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

$f'' (t) = 6 (t^{2 + 2} \cdot 4) + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.1.3

Add $2$ and $2$ .

$f'' (t) = 6 (t^{4} \cdot 4) + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.2

Move $4$ to the left of $t^{4}$ .

$f'' (t) = 6 (4 \cdot t^{4}) + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.3

Multiply $4$ by $6$ .

$f'' (t) = 24 t^{4} + 6 (t^{2} (4 \cdot - 4)) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.4

Multiply $4$ by $- 4$ .

$f'' (t) = 24 t^{4} + 6 (t^{2} \cdot - 16) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.5

Move $- 16$ to the left of $t^{2}$ .

$f'' (t) = 24 t^{4} + 6 (- 16 \cdot t^{2}) + 6 {(t^{2} - 4)}^{2}$

Step 3.11.4.6

Multiply $- 16$ by $6$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 {(t^{2} - 4)}^{2}$

Step 3.11.5

Simplify each term.

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Step 3.11.5.1

Rewrite ${(t^{2} - 4)}^{2}$ as $(t^{2} - 4) (t^{2} - 4)$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 ((t^{2} - 4) (t^{2} - 4))$

Step 3.11.5.2

Expand $(t^{2} - 4) (t^{2} - 4)$ using the FOIL Method.

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Step 3.11.5.2.1

Apply the distributive property.

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{2} (t^{2} - 4) - 4 (t^{2} - 4))$

Step 3.11.5.2.2

Apply the distributive property.

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{2} t^{2} + t^{2} \cdot - 4 - 4 (t^{2} - 4))$

Step 3.11.5.2.3

Apply the distributive property.

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{2} t^{2} + t^{2} \cdot - 4 - 4 t^{2} - 4 \cdot - 4)$

Step 3.11.5.3

Simplify and combine like terms.

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Step 3.11.5.3.1

Simplify each term.

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Step 3.11.5.3.1.1

Multiply $t^{2}$ by $t^{2}$ by adding the exponents.

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Step 3.11.5.3.1.1.1

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

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{2 + 2} + t^{2} \cdot - 4 - 4 t^{2} - 4 \cdot - 4)$

Step 3.11.5.3.1.1.2

Add $2$ and $2$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{4} + t^{2} \cdot - 4 - 4 t^{2} - 4 \cdot - 4)$

Step 3.11.5.3.1.2

Move $- 4$ to the left of $t^{2}$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{4} - 4 \cdot t^{2} - 4 t^{2} - 4 \cdot - 4)$

Step 3.11.5.3.1.3

Multiply $- 4$ by $- 4$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{4} - 4 t^{2} - 4 t^{2} + 16)$

Step 3.11.5.3.2

Subtract $4 t^{2}$ from $- 4 t^{2}$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 (t^{4} - 8 t^{2} + 16)$

Step 3.11.5.4

Apply the distributive property.

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 t^{4} + 6 (- 8 t^{2}) + 6 \cdot 16$

Step 3.11.5.5

Simplify.

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Step 3.11.5.5.1

Multiply $- 8$ by $6$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 t^{4} - 48 t^{2} + 6 \cdot 16$

Step 3.11.5.5.2

Multiply $6$ by $16$ .

$f'' (t) = 24 t^{4} - 96 t^{2} + 6 t^{4} - 48 t^{2} + 96$

Step 3.11.6

Add $24 t^{4}$ and $6 t^{4}$ .

$f'' (t) = 30 t^{4} - 96 t^{2} - 48 t^{2} + 96$

Step 3.11.7

Subtract $48 t^{2}$ from $- 96 t^{2}$ .

$f'' (t) = 30 t^{4} - 144 t^{2} + 96$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$6 t {(t^{2} - 4)}^{2} = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{3}$ and $g (t) = t^{2} - 4$ .

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Step 5.1.1.1

To apply the Chain Rule, set $u$ as $t^{2} - 4$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d t} [t^{2} - 4]$

Step 5.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$3 u^{2} \frac{d}{d t} [t^{2} - 4]$

Step 5.1.1.3

Replace all occurrences of $u$ with $t^{2} - 4$ .

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

Step 5.1.2

Differentiate.

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Step 5.1.2.1

By the Sum Rule, the derivative of $t^{2} - 4$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [- 4]$ .

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Since $- 4$ is constant with respect to $t$ , the derivative of $- 4$ with respect to $t$ is $0$ .

$3 {(t^{2} - 4)}^{2} (2 t + 0)$

Step 5.1.2.4

Simplify the expression.

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Step 5.1.2.4.1

Add $2 t$ and $0$ .

$3 {(t^{2} - 4)}^{2} (2 t)$

Step 5.1.2.4.2

Multiply $2$ by $3$ .

$6 {(t^{2} - 4)}^{2} t$

Step 5.1.2.4.3

Reorder the factors of $6 {(t^{2} - 4)}^{2} t$ .

$f' (t) = 6 t {(t^{2} - 4)}^{2}$

Step 5.2

The first derivative of $f (t)$ with respect to $t$ is $6 t {(t^{2} - 4)}^{2}$ .

$6 t {(t^{2} - 4)}^{2}$

Step 6

Set the first derivative equal to $0$ then solve the equation $6 t {(t^{2} - 4)}^{2} = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$6 t {(t^{2} - 4)}^{2} = 0$

Step 6.2

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

$t = 0$

${(t^{2} - 4)}^{2} = 0$

Step 6.3

Set $t$ equal to $0$ .

$t = 0$

Step 6.4

Set ${(t^{2} - 4)}^{2}$ equal to $0$ and solve for $t$ .

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Step 6.4.1

Set ${(t^{2} - 4)}^{2}$ equal to $0$ .

${(t^{2} - 4)}^{2} = 0$

Step 6.4.2

Solve ${(t^{2} - 4)}^{2} = 0$ for $t$ .

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Step 6.4.2.1

Factor the left side of the equation.

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Step 6.4.2.1.1

Rewrite $4$ as $2^{2}$ .

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

Step 6.4.2.1.2

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

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

Step 6.4.2.1.3

Apply the product rule to $(t + 2) (t - 2)$ .

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

Step 6.4.2.2

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

${(t + 2)}^{2} = 0$

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

Step 6.4.2.3

Set ${(t + 2)}^{2}$ equal to $0$ and solve for $t$ .

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Step 6.4.2.3.1

Set ${(t + 2)}^{2}$ equal to $0$ .

${(t + 2)}^{2} = 0$

Step 6.4.2.3.2

Solve ${(t + 2)}^{2} = 0$ for $t$ .

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Step 6.4.2.3.2.1

Set the $t + 2$ equal to $0$ .

$t + 2 = 0$

Step 6.4.2.3.2.2

Subtract $2$ from both sides of the equation.

$t = - 2$

Step 6.4.2.4

Set ${(t - 2)}^{2}$ equal to $0$ and solve for $t$ .

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Step 6.4.2.4.1

Set ${(t - 2)}^{2}$ equal to $0$ .

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

Step 6.4.2.4.2

Solve ${(t - 2)}^{2} = 0$ for $t$ .

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Step 6.4.2.4.2.1

Set the $t - 2$ equal to $0$ .

$t - 2 = 0$

Step 6.4.2.4.2.2

Add $2$ to both sides of the equation.

$t = 2$

Step 6.4.2.5

The final solution is all the values that make ${(t + 2)}^{2} {(t - 2)}^{2} = 0$ true.

$t = - 2, 2$

Step 6.5

The final solution is all the values that make $6 t {(t^{2} - 4)}^{2} = 0$ true.

$t = 0, - 2, 2$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 8

Critical points to evaluate.

$t = 0, - 2, 2$

Step 9

Evaluate the second derivative at $t = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$30 {(0)}^{4} - 144 {(0)}^{2} + 96$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

$30 \cdot 0 - 144 {(0)}^{2} + 96$

Step 10.1.2

Multiply $30$ by $0$ .

$0 - 144 {(0)}^{2} + 96$

Step 10.1.3

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

$0 - 144 \cdot 0 + 96$

Step 10.1.4

Multiply $- 144$ by $0$ .

$0 + 0 + 96$

Step 10.2

Simplify by adding numbers.

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Step 10.2.1

Add $0$ and $0$ .

$0 + 96$

Step 10.2.2

Add $0$ and $96$ .

$96$

Step 11

$t = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$t = 0$ is a local minimum

Step 12

Find the y-value when $t = 0$ .

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Step 12.1

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

$f (0) = {({(0)}^{2} - 4)}^{3}$

Step 12.2

Simplify the result.

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Step 12.2.1

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

$f (0) = {(0 - 4)}^{3}$

Step 12.2.2

Subtract $4$ from $0$ .

$f (0) = {(- 4)}^{3}$

Step 12.2.3

Raise $- 4$ to the power of $3$ .

$f (0) = - 64$

Step 12.2.4

The final answer is $- 64$ .

$y = - 64$

Step 13

Evaluate the second derivative at $t = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$30 {(- 2)}^{4} - 144 {(- 2)}^{2} + 96$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

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

$30 \cdot 16 - 144 {(- 2)}^{2} + 96$

Step 14.1.2

Multiply $30$ by $16$ .

$480 - 144 {(- 2)}^{2} + 96$

Step 14.1.3

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

$480 - 144 \cdot 4 + 96$

Step 14.1.4

Multiply $- 144$ by $4$ .

$480 - 576 + 96$

Step 14.2

Simplify by adding and subtracting.

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Step 14.2.1

Subtract $576$ from $480$ .

$- 96 + 96$

Step 14.2.2

Add $- 96$ and $96$ .

$0$

Step 15

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 15.1

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

$(- \infty, - 2) \cup (- 2, 0) \cup (0, 2) \cup (2, \infty)$

Step 15.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 2)$ in the first derivative $6 t {(t^{2} - 4)}^{2}$ to check if the result is negative or positive.

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Step 15.2.1

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

$f' (- 4) = 6 (- 4) {({(- 4)}^{2} - 4)}^{2}$

Step 15.2.2

Simplify the result.

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Step 15.2.2.1

Multiply $6$ by $- 4$ .

$f' (- 4) = - 24 {({(- 4)}^{2} - 4)}^{2}$

Step 15.2.2.2

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

$f' (- 4) = - 24 {(16 - 4)}^{2}$

Step 15.2.2.3

Subtract $4$ from $16$ .

$f' (- 4) = - 24 \cdot 12^{2}$

Step 15.2.2.4

Raise $12$ to the power of $2$ .

$f' (- 4) = - 24 \cdot 144$

Step 15.2.2.5

Multiply $- 24$ by $144$ .

$f' (- 4) = - 3456$

Step 15.2.2.6

The final answer is $- 3456$ .

$- 3456$

Step 15.3

Substitute any number, such as $- 1$ , from the interval $(- 2, 0)$ in the first derivative $6 t {(t^{2} - 4)}^{2}$ to check if the result is negative or positive.

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Step 15.3.1

Replace the variable $t$ with $- 1$ in the expression.

$f' (- 1) = 6 (- 1) {({(- 1)}^{2} - 4)}^{2}$

Step 15.3.2

Simplify the result.

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Step 15.3.2.1

Multiply $6$ by $- 1$ .

$f' (- 1) = - 6 {({(- 1)}^{2} - 4)}^{2}$

Step 15.3.2.2

Raise $- 1$ to the power of $2$ .

$f' (- 1) = - 6 {(1 - 4)}^{2}$

Step 15.3.2.3

Subtract $4$ from $1$ .

$f' (- 1) = - 6 {(- 3)}^{2}$

Step 15.3.2.4

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

$f' (- 1) = - 6 \cdot 9$

Step 15.3.2.5

Multiply $- 6$ by $9$ .

$f' (- 1) = - 54$

Step 15.3.2.6

The final answer is $- 54$ .

$- 54$

Step 15.4

Substitute any number, such as $1$ , from the interval $(0, 2)$ in the first derivative $6 t {(t^{2} - 4)}^{2}$ to check if the result is negative or positive.

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Step 15.4.1

Replace the variable $t$ with $1$ in the expression.

$f' (1) = 6 (1) {({(1)}^{2} - 4)}^{2}$

Step 15.4.2

Simplify the result.

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Step 15.4.2.1

Multiply $6$ by $1$ .

$f' (1) = 6 {({(1)}^{2} - 4)}^{2}$

Step 15.4.2.2

One to any power is one.

$f' (1) = 6 {(1 - 4)}^{2}$

Step 15.4.2.3

Subtract $4$ from $1$ .

$f' (1) = 6 {(- 3)}^{2}$

Step 15.4.2.4

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

$f' (1) = 6 \cdot 9$

Step 15.4.2.5

Multiply $6$ by $9$ .

$f' (1) = 54$

Step 15.4.2.6

The final answer is $54$ .

$54$

Step 15.5

Substitute any number, such as $4$ , from the interval $(2, \infty)$ in the first derivative $6 t {(t^{2} - 4)}^{2}$ to check if the result is negative or positive.

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Step 15.5.1

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

$f' (4) = 6 (4) {({(4)}^{2} - 4)}^{2}$

Step 15.5.2

Simplify the result.

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Step 15.5.2.1

Multiply $6$ by $4$ .

$f' (4) = 24 {({(4)}^{2} - 4)}^{2}$

Step 15.5.2.2

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

$f' (4) = 24 {(16 - 4)}^{2}$

Step 15.5.2.3

Subtract $4$ from $16$ .

$f' (4) = 24 \cdot 12^{2}$

Step 15.5.2.4

Raise $12$ to the power of $2$ .

$f' (4) = 24 \cdot 144$

Step 15.5.2.5

Multiply $24$ by $144$ .

$f' (4) = 3456$

Step 15.5.2.6

The final answer is $3456$ .

$3456$

Step 15.6

Since the first derivative did not change signs around $t = - 2$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 15.7

Since the first derivative changed signs from negative to positive around $t = 0$ , then $t = 0$ is a local minimum.

$t = 0$ is a local minimum

Step 15.8

Since the first derivative did not change signs around $t = 2$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 15.9

These are the local extrema for $f (t) = {(t^{2} - 4)}^{3}$ .

$t = 0$ is a local minimum

Step 16