Find the Critical Points (x^2-1)^3

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

Find the derivative.

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Simplify the expression.

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Add $2 x$ and $0$ .

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

Multiply $2$ by $3$ .

$6 {(x^{2} - 1)}^{2} x$

Reorder the factors of $6 {(x^{2} - 1)}^{2} x$ .

$6 x {(x^{2} - 1)}^{2}$

Set the derivative equal to $0$ .

$6 x {(x^{2} - 1)}^{2} = 0$

Solve for $x$ .

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Divide each term in $6 x {(x^{2} - 1)}^{2} = 0$ by $6$ .

$\frac{6 x {(x^{2} - 1)}^{2}}{6} = \frac{0}{6}$

Cancel the common factor of $6$ .

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Cancel the common factor.

$\frac{6 x {(x^{2} - 1)}^{2}}{6} = \frac{0}{6}$

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

$x {(x^{2} - 1)}^{2} = \frac{0}{6}$

Divide $0$ by $6$ .

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

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

$x = 0$

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

Set the first factor equal to $0$ .

$x = 0$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

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

Factor the left side of the equation.

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

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

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 $b = 1$ .

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

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

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

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

${(x + 1)}^{2} = 0$

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

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

${(x + 1)}^{2} = 0$

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 1, 1$

The final solution is all the values that make $\frac{6 x {(x^{2} - 1)}^{2}}{6} = \frac{0}{6}$ true.

$x = 0, - 1, 1$

Substitute the values of $x$ which cause the derivative to be $0$ into the original function.

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

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