Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x^{2} - 1$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 1$ .

$\frac{d}{d u} [u^{4}] \frac{d}{d x} [x^{2} - 1]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$4 u^{3} \frac{d}{d x} [x^{2} - 1]$

Step 1.1.3

Replace all occurrences of $u$ with $x^{2} - 1$ .

$4 {(x^{2} - 1)}^{3} \frac{d}{d x} [x^{2} - 1]$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$ .

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

Step 1.2.2

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

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

Step 1.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$4 {(x^{2} - 1)}^{3} (2 x + 0)$

Step 1.2.4

Simplify the expression.

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Step 1.2.4.1

Add $2 x$ and $0$ .

$4 {(x^{2} - 1)}^{3} (2 x)$

Step 1.2.4.2

Multiply $2$ by $4$ .

$8 {(x^{2} - 1)}^{3} x$

Step 1.2.4.3

Reorder the factors of $8 {(x^{2} - 1)}^{3} x$ .

$8 x {(x^{2} - 1)}^{3}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x {(x^{2} - 1)}^{3}$ with respect to $x$ is $8 \frac{d}{d x} [x {(x^{2} - 1)}^{3}]$ .

$f'' (x) = 8 \frac{d}{d x} (x {(x^{2} - 1)}^{3})$

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = {(x^{2} - 1)}^{3}$ .

$f'' (x) = 8 (x \frac{d}{d x} ({(x^{2} - 1)}^{3}) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x^{2} - 1$ .

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Step 2.3.1

To apply the Chain Rule, set $u$ as $x^{2} - 1$ .

$f'' (x) = 8 (x (\frac{d}{d u} (u^{3}) \frac{d}{d x} (x^{2} - 1)) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.3.2

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

$f'' (x) = 8 (x (3 u^{2} \frac{d}{d x} (x^{2} - 1)) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.3.3

Replace all occurrences of $u$ with $x^{2} - 1$ .

$f'' (x) = 8 (x (3 {(x^{2} - 1)}^{2} \frac{d}{d x} (x^{2} - 1)) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.4

Differentiate.

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Step 2.4.1

By the Sum Rule, the derivative of $x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = 8 (x (3 {(x^{2} - 1)}^{2} (\frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 1))) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.4.2

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

$f'' (x) = 8 (x (3 {(x^{2} - 1)}^{2} (2 x + \frac{d}{d x} (- 1))) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.4.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = 8 (x (3 {(x^{2} - 1)}^{2} (2 x + 0)) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.4.4

Simplify the expression.

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Step 2.4.4.1

Add $2 x$ and $0$ .

$f'' (x) = 8 (x (3 {(x^{2} - 1)}^{2} (2 x)) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.4.4.2

Multiply $2$ by $3$ .

$f'' (x) = 8 (x (6 {(x^{2} - 1)}^{2} x) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.5

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

$f'' (x) = 8 (x \cdot x (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.6

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

$f'' (x) = 8 (x \cdot x (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.7

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

$f'' (x) = 8 (x^{1 + 1} (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.8

Add $1$ and $1$ .

$f'' (x) = 8 (x^{2} (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3} \frac{d}{d x} (x))$

Step 2.9

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

$f'' (x) = 8 (x^{2} (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3} \cdot 1)$

Step 2.10

Multiply ${(x^{2} - 1)}^{3}$ by $1$ .

$f'' (x) = 8 (x^{2} (6 {(x^{2} - 1)}^{2}) + {(x^{2} - 1)}^{3})$

Step 2.11

Simplify.

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Step 2.11.1

Apply the distributive property.

$f'' (x) = 8 (x^{2} (6 {(x^{2} - 1)}^{2})) + 8 {(x^{2} - 1)}^{3}$

Step 2.11.2

Multiply $6$ by $8$ .

$f'' (x) = 48 (x^{2} ({(x^{2} - 1)}^{2})) + 8 {(x^{2} - 1)}^{3}$

Step 2.11.3

Factor $8 {(x^{2} - 1)}^{2}$ out of $48 x^{2} {(x^{2} - 1)}^{2} + 8 {(x^{2} - 1)}^{3}$ .

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Step 2.11.3.1

Factor $8 {(x^{2} - 1)}^{2}$ out of $48 x^{2} {(x^{2} - 1)}^{2}$ .

$f'' (x) = 8 {(x^{2} - 1)}^{2} (6 x^{2}) + 8 {(x^{2} - 1)}^{3}$

Step 2.11.3.2

Factor $8 {(x^{2} - 1)}^{2}$ out of $8 {(x^{2} - 1)}^{3}$ .

$f'' (x) = 8 {(x^{2} - 1)}^{2} (6 x^{2}) + 8 {(x^{2} - 1)}^{2} (x^{2} - 1)$

Step 2.11.3.3

Factor $8 {(x^{2} - 1)}^{2}$ out of $8 {(x^{2} - 1)}^{2} (6 x^{2}) + 8 {(x^{2} - 1)}^{2} (x^{2} - 1)$ .

$f'' (x) = 8 {(x^{2} - 1)}^{2} (6 x^{2} + x^{2} - 1)$

Step 2.11.4

Add $6 x^{2}$ and $x^{2}$ .

$f'' (x) = 8 {(x^{2} - 1)}^{2} (7 x^{2} - 1)$

Step 2.11.5

Rewrite ${(x^{2} - 1)}^{2}$ as $(x^{2} - 1) (x^{2} - 1)$ .

$f'' (x) = 8 ((x^{2} - 1) (x^{2} - 1)) (7 x^{2} - 1)$

Step 2.11.6

Expand $(x^{2} - 1) (x^{2} - 1)$ using the FOIL Method.

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Step 2.11.6.1

Apply the distributive property.

$f'' (x) = 8 (x^{2} (x^{2} - 1) - 1 (x^{2} - 1)) (7 x^{2} - 1)$

Step 2.11.6.2

Apply the distributive property.

$f'' (x) = 8 (x^{2} x^{2} + x^{2} \cdot - 1 - 1 (x^{2} - 1)) (7 x^{2} - 1)$

Step 2.11.6.3

Apply the distributive property.

$f'' (x) = 8 (x^{2} x^{2} + x^{2} \cdot - 1 - 1 x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7

Simplify and combine like terms.

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Step 2.11.7.1

Simplify each term.

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Step 2.11.7.1.1

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

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Step 2.11.7.1.1.1

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

$f'' (x) = 8 (x^{2 + 2} + x^{2} \cdot - 1 - 1 x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7.1.1.2

Add $2$ and $2$ .

$f'' (x) = 8 (x^{4} + x^{2} \cdot - 1 - 1 x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7.1.2

Move $- 1$ to the left of $x^{2}$ .

$f'' (x) = 8 (x^{4} - 1 \cdot x^{2} - 1 x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7.1.3

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$f'' (x) = 8 (x^{4} - x^{2} - 1 x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7.1.4

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$f'' (x) = 8 (x^{4} - x^{2} - x^{2} - 1 \cdot - 1) (7 x^{2} - 1)$

Step 2.11.7.1.5

Multiply $- 1$ by $- 1$ .

$f'' (x) = 8 (x^{4} - x^{2} - x^{2} + 1) (7 x^{2} - 1)$

Step 2.11.7.2

Subtract $x^{2}$ from $- x^{2}$ .

$f'' (x) = 8 (x^{4} - 2 x^{2} + 1) (7 x^{2} - 1)$

Step 2.11.8

Apply the distributive property.

$f'' (x) = (8 x^{4} + 8 (- 2 x^{2}) + 8 \cdot 1) (7 x^{2} - 1)$

Step 2.11.9

Simplify.

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Step 2.11.9.1

Multiply $- 2$ by $8$ .

$f'' (x) = (8 x^{4} - 16 x^{2} + 8 \cdot 1) (7 x^{2} - 1)$

Step 2.11.9.2

Multiply $8$ by $1$ .

$f'' (x) = (8 x^{4} - 16 x^{2} + 8) (7 x^{2} - 1)$

Step 2.11.10

Expand $(8 x^{4} - 16 x^{2} + 8) (7 x^{2} - 1)$ by multiplying each term in the first expression by each term in the second expression.

$f'' (x) = 8 x^{4} (7 x^{2}) + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11

Simplify each term.

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Step 2.11.11.1

Rewrite using the commutative property of multiplication.

$f'' (x) = 8 \cdot (7 x^{4} x^{2}) + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.2

Multiply $x^{4}$ by $x^{2}$ by adding the exponents.

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Step 2.11.11.2.1

Move $x^{2}$ .

$f'' (x) = 8 \cdot (7 (x^{2} x^{4})) + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.2.2

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

$f'' (x) = 8 \cdot (7 x^{2 + 4}) + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.2.3

Add $2$ and $4$ .

$f'' (x) = 8 \cdot (7 x^{6}) + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.3

Multiply $8$ by $7$ .

$f'' (x) = 56 x^{6} + 8 x^{4} \cdot - 1 - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.4

Multiply $- 1$ by $8$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 16 x^{2} (7 x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.5

Rewrite using the commutative property of multiplication.

$f'' (x) = 56 x^{6} - 8 x^{4} - 16 \cdot (7 x^{2} x^{2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.6

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

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Step 2.11.11.6.1

Move $x^{2}$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 16 \cdot (7 (x^{2} x^{2})) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.6.2

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

$f'' (x) = 56 x^{6} - 8 x^{4} - 16 \cdot (7 x^{2 + 2}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.6.3

Add $2$ and $2$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 16 \cdot (7 x^{4}) - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.7

Multiply $- 16$ by $7$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 112 x^{4} - 16 x^{2} \cdot - 1 + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.8

Multiply $- 1$ by $- 16$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 112 x^{4} + 16 x^{2} + 8 (7 x^{2}) + 8 \cdot - 1$

Step 2.11.11.9

Multiply $7$ by $8$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 112 x^{4} + 16 x^{2} + 56 x^{2} + 8 \cdot - 1$

Step 2.11.11.10

Multiply $8$ by $- 1$ .

$f'' (x) = 56 x^{6} - 8 x^{4} - 112 x^{4} + 16 x^{2} + 56 x^{2} - 8$

Step 2.11.12

Subtract $112 x^{4}$ from $- 8 x^{4}$ .

$f'' (x) = 56 x^{6} - 120 x^{4} + 16 x^{2} + 56 x^{2} - 8$

Step 2.11.13

Add $16 x^{2}$ and $56 x^{2}$ .

$f'' (x) = 56 x^{6} - 120 x^{4} + 72 x^{2} - 8$

Step 3

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

$8 x {(x^{2} - 1)}^{3} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x^{2} - 1$ .

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Step 4.1.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 1$ .

$f' (x) = \frac{d}{d u} (u^{4}) \frac{d}{d x} (x^{2} - 1)$

Step 4.1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

$f' (x) = 4 u^{3} \frac{d}{d x} (x^{2} - 1)$

Step 4.1.1.3

Replace all occurrences of $u$ with $x^{2} - 1$ .

$f' (x) = 4 {(x^{2} - 1)}^{3} \frac{d}{d x} (x^{2} - 1)$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$ .

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

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

Step 4.1.2.4

Simplify the expression.

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Step 4.1.2.4.1

Add $2 x$ and $0$ .

$f' (x) = 4 {(x^{2} - 1)}^{3} (2 x)$

Step 4.1.2.4.2

Multiply $2$ by $4$ .

$f' (x) = 8 {(x^{2} - 1)}^{3} x$

Step 4.1.2.4.3

Reorder the factors of $8 {(x^{2} - 1)}^{3} x$ .

$f' (x) = 8 x {(x^{2} - 1)}^{3}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $8 x {(x^{2} - 1)}^{3}$ .

$8 x {(x^{2} - 1)}^{3}$

Step 5

Set the first derivative equal to $0$ then solve the equation $8 x {(x^{2} - 1)}^{3} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$8 x {(x^{2} - 1)}^{3} = 0$

Step 5.2

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

$x = 0$

${(x^{2} - 1)}^{3} = 0$

Step 5.3

Set $x$ equal to $0$ .

$x = 0$

Step 5.4

Set ${(x^{2} - 1)}^{3}$ equal to $0$ and solve for $x$ .

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Step 5.4.1

Set ${(x^{2} - 1)}^{3}$ equal to $0$ .

${(x^{2} - 1)}^{3} = 0$

Step 5.4.2

Solve ${(x^{2} - 1)}^{3} = 0$ for $x$ .

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Step 5.4.2.1

Factor the left side of the equation.

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Step 5.4.2.1.1

Rewrite $1$ as $1^{2}$ .

${(x^{2} - 1^{2})}^{3} = 0$

Step 5.4.2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

${((x + 1) (x - 1))}^{3} = 0$

Step 5.4.2.1.3

Apply the product rule to $(x + 1) (x - 1)$ .

${(x + 1)}^{3} {(x - 1)}^{3} = 0$

Step 5.4.2.2

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

${(x + 1)}^{3} = 0$

${(x - 1)}^{3} = 0$

Step 5.4.2.3

Set ${(x + 1)}^{3}$ equal to $0$ and solve for $x$ .

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Step 5.4.2.3.1

Set ${(x + 1)}^{3}$ equal to $0$ .

${(x + 1)}^{3} = 0$

Step 5.4.2.3.2

Solve ${(x + 1)}^{3} = 0$ for $x$ .

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Step 5.4.2.3.2.1

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 5.4.2.3.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.4.2.4

Set ${(x - 1)}^{3}$ equal to $0$ and solve for $x$ .

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Step 5.4.2.4.1

Set ${(x - 1)}^{3}$ equal to $0$ .

${(x - 1)}^{3} = 0$

Step 5.4.2.4.2

Solve ${(x - 1)}^{3} = 0$ for $x$ .

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Step 5.4.2.4.2.1

Set the $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 5.4.2.4.2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5.4.2.5

The final solution is all the values that make ${(x + 1)}^{3} {(x - 1)}^{3} = 0$ true.

$x = - 1, 1$

Step 5.5

The final solution is all the values that make $8 x {(x^{2} - 1)}^{3} = 0$ true.

$x = 0, - 1, 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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 7

Critical points to evaluate.

$x = 0, - 1, 1$

Step 8

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

$56 {(0)}^{6} - 120 {(0)}^{4} + 72 {(0)}^{2} - 8$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

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

$56 \cdot 0 - 120 {(0)}^{4} + 72 {(0)}^{2} - 8$

Step 9.1.2

Multiply $56$ by $0$ .

$0 - 120 {(0)}^{4} + 72 {(0)}^{2} - 8$

Step 9.1.3

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

$0 - 120 \cdot 0 + 72 {(0)}^{2} - 8$

Step 9.1.4

Multiply $- 120$ by $0$ .

$0 + 0 + 72 {(0)}^{2} - 8$

Step 9.1.5

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

$0 + 0 + 72 \cdot 0 - 8$

Step 9.1.6

Multiply $72$ by $0$ .

$0 + 0 + 0 - 8$

Step 9.2

Simplify by adding and subtracting.

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Step 9.2.1

Add $0$ and $0$ .

$0 + 0 - 8$

Step 9.2.2

Add $0$ and $0$ .

$0 - 8$

Step 9.2.3

Subtract $8$ from $0$ .

$- 8$

Step 10

$x = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 0$ is a local maximum

Step 11

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

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Step 11.1

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

$f (0) = {({(0)}^{2} - 1)}^{4}$

Step 11.2

Simplify the result.

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Step 11.2.1

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

$f (0) = {(0 - 1)}^{4}$

Step 11.2.2

Subtract $1$ from $0$ .

$f (0) = {(- 1)}^{4}$

Step 11.2.3

Raise $- 1$ to the power of $4$ .

$f (0) = 1$

Step 11.2.4

The final answer is $1$ .

$y = 1$

Step 12

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$56 {(- 1)}^{6} - 120 {(- 1)}^{4} + 72 {(- 1)}^{2} - 8$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Raise $- 1$ to the power of $6$ .

$56 \cdot 1 - 120 {(- 1)}^{4} + 72 {(- 1)}^{2} - 8$

Step 13.1.2

Multiply $56$ by $1$ .

$56 - 120 {(- 1)}^{4} + 72 {(- 1)}^{2} - 8$

Step 13.1.3

Raise $- 1$ to the power of $4$ .

$56 - 120 \cdot 1 + 72 {(- 1)}^{2} - 8$

Step 13.1.4

Multiply $- 120$ by $1$ .

$56 - 120 + 72 {(- 1)}^{2} - 8$

Step 13.1.5

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

$56 - 120 + 72 \cdot 1 - 8$

Step 13.1.6

Multiply $72$ by $1$ .

$56 - 120 + 72 - 8$

Step 13.2

Simplify by adding and subtracting.

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Step 13.2.1

Subtract $120$ from $56$ .

$- 64 + 72 - 8$

Step 13.2.2

Add $- 64$ and $72$ .

$8 - 8$

Step 13.2.3

Subtract $8$ from $8$ .

$0$

Step 14

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

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Step 14.1

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

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

Step 14.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - 1)$ in the first derivative $8 x {(x^{2} - 1)}^{3}$ to check if the result is negative or positive.

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Step 14.2.1

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

$f' (- 4) = 8 (- 4) {({(- 4)}^{2} - 1)}^{3}$

Step 14.2.2

Simplify the result.

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Step 14.2.2.1

Multiply $8$ by $- 4$ .

$f' (- 4) = - 32 {({(- 4)}^{2} - 1)}^{3}$

Step 14.2.2.2

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

$f' (- 4) = - 32 {(16 - 1)}^{3}$

Step 14.2.2.3

Subtract $1$ from $16$ .

$f' (- 4) = - 32 \cdot 15^{3}$

Step 14.2.2.4

Raise $15$ to the power of $3$ .

$f' (- 4) = - 32 \cdot 3375$

Step 14.2.2.5

Multiply $- 32$ by $3375$ .

$f' (- 4) = - 108000$

Step 14.2.2.6

The final answer is $- 108000$ .

$- 108000$

Step 14.3

Substitute any number, such as $- 0.5$ , from the interval $(- 1, 0)$ in the first derivative $8 x {(x^{2} - 1)}^{3}$ to check if the result is negative or positive.

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Step 14.3.1

Replace the variable $x$ with $- 0.5$ in the expression.

$f' (- 0.5) = 8 (- 0.5) {({(- 0.5)}^{2} - 1)}^{3}$

Step 14.3.2

Simplify the result.

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Step 14.3.2.1

Multiply $8$ by $- 0.5$ .

$f' (- 0.5) = - 4 {({(- 0.5)}^{2} - 1)}^{3}$

Step 14.3.2.2

Raise $- 0.5$ to the power of $2$ .

$f' (- 0.5) = - 4 {(0.25 - 1)}^{3}$

Step 14.3.2.3

Subtract $1$ from $0.25$ .

$f' (- 0.5) = - 4 {(- 0.75)}^{3}$

Step 14.3.2.4

Raise $- 0.75$ to the power of $3$ .

$f' (- 0.5) = - 4 \cdot - 0.421875$

Step 14.3.2.5

Multiply $- 4$ by $- 0.421875$ .

$f' (- 0.5) = 1.6875$

Step 14.3.2.6

The final answer is $1.6875$ .

$1.6875$

Step 14.4

Substitute any number, such as $0.5$ , from the interval $(0, 1)$ in the first derivative $8 x {(x^{2} - 1)}^{3}$ to check if the result is negative or positive.

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Step 14.4.1

Replace the variable $x$ with $0.5$ in the expression.

$f' (0.5) = 8 (0.5) {({(0.5)}^{2} - 1)}^{3}$

Step 14.4.2

Simplify the result.

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Step 14.4.2.1

Multiply $8$ by $0.5$ .

$f' (0.5) = 4 {({(0.5)}^{2} - 1)}^{3}$

Step 14.4.2.2

Raise $0.5$ to the power of $2$ .

$f' (0.5) = 4 {(0.25 - 1)}^{3}$

Step 14.4.2.3

Subtract $1$ from $0.25$ .

$f' (0.5) = 4 {(- 0.75)}^{3}$

Step 14.4.2.4

Raise $- 0.75$ to the power of $3$ .

$f' (0.5) = 4 \cdot - 0.421875$

Step 14.4.2.5

Multiply $4$ by $- 0.421875$ .

$f' (0.5) = - 1.6875$

Step 14.4.2.6

The final answer is $- 1.6875$ .

$- 1.6875$

Step 14.5

Substitute any number, such as $4$ , from the interval $(1, \infty)$ in the first derivative $8 x {(x^{2} - 1)}^{3}$ to check if the result is negative or positive.

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Step 14.5.1

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

$f' (4) = 8 (4) {({(4)}^{2} - 1)}^{3}$

Step 14.5.2

Simplify the result.

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Step 14.5.2.1

Multiply $8$ by $4$ .

$f' (4) = 32 {({(4)}^{2} - 1)}^{3}$

Step 14.5.2.2

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

$f' (4) = 32 {(16 - 1)}^{3}$

Step 14.5.2.3

Subtract $1$ from $16$ .

$f' (4) = 32 \cdot 15^{3}$

Step 14.5.2.4

Raise $15$ to the power of $3$ .

$f' (4) = 32 \cdot 3375$

Step 14.5.2.5

Multiply $32$ by $3375$ .

$f' (4) = 108000$

Step 14.5.2.6

The final answer is $108000$ .

$108000$

Step 14.6

Since the first derivative changed signs from negative to positive around $x = - 1$ , then $x = - 1$ is a local minimum.

$x = - 1$ is a local minimum

Step 14.7

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

$x = 0$ is a local maximum

Step 14.8

Since the first derivative changed signs from negative to positive around $x = 1$ , then $x = 1$ is a local minimum.

$x = 1$ is a local minimum

Step 14.9

These are the local extrema for $f (x) = {(x^{2} - 1)}^{4}$ .

$x = - 1$ is a local minimum

$x = 0$ is a local maximum

$x = 1$ is a local minimum

$x = - 1$ is a local minimum

$x = 0$ is a local maximum

$x = 1$ is a local minimum

Step 15