Find the Local Maxima and Minima f(x)=x^4-8x^2+1

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

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x^{4} - 8 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 8 x^{2}] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 8 x^{2}] + \frac{d}{d x} [1]$

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

$4 x^{3} + \frac{d}{d x} [- 8 x^{2}] + \frac{d}{d x} [1]$

Evaluate $\frac{d}{d x} [- 8 x^{2}]$ .

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

$4 x^{3} - 8 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$

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

$4 x^{3} - 8 (2 x) + \frac{d}{d x} [1]$

Multiply $2$ by $- 8$ .

$4 x^{3} - 16 x + \frac{d}{d x} [1]$

Differentiate using the Constant Rule.

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

$4 x^{3} - 16 x + 0$

Add $4 x^{3} - 16 x$ and $0$ .

$4 x^{3} - 16 x$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $4 x^{3} - 16 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 16 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 16 x)$

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 16 x)$

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 16 x)$

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 16 x)$

Evaluate $\frac{d}{d x} [- 16 x]$ .

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Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 x$ with respect to $x$ is $- 16 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 16 \frac{d}{d x} x$

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

$f'' (x) = 12 x^{2} - 16 \cdot 1$

Multiply $- 16$ by $1$ .

$f'' (x) = 12 x^{2} - 16$

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

$4 x^{3} - 16 x = 0$

Factor the left side of the equation.

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Factor $4 x$ out of $4 x^{3} - 16 x$ .

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Factor $4 x$ out of $4 x^{3}$ .

$4 x (x^{2}) - 16 x = 0$

Factor $4 x$ out of $- 16 x$ .

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

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 4)$ .

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

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

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

Factor.

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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 $b = 2$ .

$4 x ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$4 x (x + 2) (x - 2) = 0$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Divide $0$ by $4$ .

$x = 0$

Set $x + 2$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x - 2$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The solution is the result of $4 x = 0, x + 2 = 0,$ and $x - 2 = 0$ .

$x = 0, - 2, 2$

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.

$12 {(0)}^{2} - 16$

Evaluate the second derivative.

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Simplify each term.

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

$12 \cdot 0 - 16$

Multiply $12$ by $0$ .

$0 - 16$

Subtract $16$ from $0$ .

$- 16$

$x = 0$ 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

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

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

$f (0) = {(0)}^{4} - 8 {(0)}^{2} + 1$

Simplify the result.

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Simplify each term.

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

$f (0) = 0 - 8 {(0)}^{2} + 1$

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

$f (0) = 0 - 8 \cdot 0 + 1$

Multiply $- 8$ by $0$ .

$f (0) = 0 + 0 + 1$

Simplify by adding zeros.

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Add $0$ and $0$ .

$f (0) = 0 + 1$

Add $0$ and $1$ .

$f (0) = 1$

The final answer is $1$ .

$y = 1$

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

$12 {(- 2)}^{2} - 16$

Evaluate the second derivative.

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Simplify each term.

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

$12 \cdot 4 - 16$

Multiply $12$ by $4$ .

$48 - 16$

Subtract $16$ from $48$ .

$32$

$x = - 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 2$ is a local minimum

Find the y-value when $x = - 2$ .

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Replace the variable $x$ with $- 2$ in the expression.

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

Simplify the result.

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Simplify each term.

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

$f (- 2) = 16 - 8 {(- 2)}^{2} + 1$

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

$f (- 2) = 16 - 8 \cdot 4 + 1$

Multiply $- 8$ by $4$ .

$f (- 2) = 16 - 32 + 1$

Simplify by adding and subtracting.

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Subtract $32$ from $16$ .

$f (- 2) = - 16 + 1$

Add $- 16$ and $1$ .

$f (- 2) = - 15$

The final answer is $- 15$ .

$y = - 15$

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

$12 {(2)}^{2} - 16$

Evaluate the second derivative.

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Simplify each term.

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

$12 \cdot 4 - 16$

Multiply $12$ by $4$ .

$48 - 16$

Subtract $16$ from $48$ .

$32$

$x = 2$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 2$ is a local minimum

Find the y-value when $x = 2$ .

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Replace the variable $x$ with $2$ in the expression.

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

Simplify the result.

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Simplify each term.

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

$f (2) = 16 - 8 {(2)}^{2} + 1$

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

$f (2) = 16 - 8 \cdot 4 + 1$

Multiply $- 8$ by $4$ .

$f (2) = 16 - 32 + 1$

Simplify by adding and subtracting.

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Subtract $32$ from $16$ .

$f (2) = - 16 + 1$

Add $- 16$ and $1$ .

$f (2) = - 15$

The final answer is $- 15$ .

$y = - 15$

These are the local extrema for $f (x) = x^{4} - 8 x^{2} + 1$ .

$(0, 1)$ is a local maxima

$(- 2, - 15)$ is a local minima

$(2, - 15)$ is a local minima

Find the Local Maxima and Minima f(x)=x^4-8x^2+1

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