Calculus Examples

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

Step 1

Find the first derivative of the function.

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Differentiate.

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

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

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

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

Evaluate $\frac{d}{d x} [- 4 x^{2}]$ .

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

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

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

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

Multiply $2$ by $- 4$ .

$4 x^{3} - 8 x$

Step 2

Find the second derivative of the function.

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By the Sum Rule, the derivative of $4 x^{3} - 8 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 8 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 8 x)$

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

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 8 x)$

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 8 x)$

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 8 x)$

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

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

$f'' (x) = 12 x^{2} - 8 \frac{d}{d x} x$

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

$f'' (x) = 12 x^{2} - 8 \cdot 1$

Multiply $- 8$ by $1$ .

$f'' (x) = 12 x^{2} - 8$

Step 3

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

$4 x^{3} - 8 x = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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Differentiate.

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

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

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

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

Evaluate $\frac{d}{d x} [- 4 x^{2}]$ .

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

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

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

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

Multiply $2$ by $- 4$ .

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

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 8 x$ .

$4 x^{3} - 8 x$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 8 x = 0$ .

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Set the first derivative equal to $0$ .

$4 x^{3} - 8 x = 0$

Factor $4 x$ out of $4 x^{3} - 8 x$ .

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

$4 x (x^{2}) - 8 x = 0$

Factor $4 x$ out of $- 8 x$ .

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

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

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

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

$x = 0$

$x^{2} - 2 = 0$

Set $x$ equal to $0$ .

$x = 0$

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

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Set $x^{2} - 2$ equal to $0$ .

$x^{2} - 2 = 0$

Solve $x^{2} - 2 = 0$ for $x$ .

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Add $2$ to both sides of the equation.

$x^{2} = 2$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{2}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{2}, - \sqrt{2}$

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

$x = 0, \sqrt{2}, - \sqrt{2}$

Step 6

Find the values where the derivative is undefined.

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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.

$x = 0, \sqrt{2}, - \sqrt{2}$

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.

$12 {(0)}^{2} - 8$

Step 9

Evaluate the second derivative.

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Simplify each term.

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

$12 \cdot 0 - 8$

Multiply $12$ by $0$ .

$0 - 8$

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

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

Simplify the result.

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Simplify each term.

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

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

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

$f (0) = 0 - 4 \cdot 0$

Multiply $- 4$ by $0$ .

$f (0) = 0 + 0$

Add $0$ and $0$ .

$f (0) = 0$

The final answer is $0$ .

$y = 0$

Step 12

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

$12 {(\sqrt{2})}^{2} - 8$

Step 13

Evaluate the second derivative.

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Simplify each term.

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Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$12 {(2^{\frac{1}{2}})}^{2} - 8$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$12 \cdot 2^{\frac{1}{2} \cdot 2} - 8$

Combine $\frac{1}{2}$ and $2$ .

$12 \cdot 2^{\frac{2}{2}} - 8$

Cancel the common factor of $2$ .

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Cancel the common factor.

$12 \cdot 2^{\frac{2}{2}} - 8$

Rewrite the expression.

$12 \cdot 2^{1} - 8$

Evaluate the exponent.

$12 \cdot 2 - 8$

Multiply $12$ by $2$ .

$24 - 8$

Subtract $8$ from $24$ .

$16$

Step 14

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

$x = \sqrt{2}$ is a local minimum

Step 15

Find the y-value when $x = \sqrt{2}$ .

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Replace the variable $x$ with $\sqrt{2}$ in the expression.

$f (\sqrt{2}) = {(\sqrt{2})}^{4} - 4 {(\sqrt{2})}^{2}$

Simplify the result.

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Simplify each term.

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

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$f (\sqrt{2}) = {(2^{\frac{1}{2}})}^{4} - 4 {(\sqrt{2})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\sqrt{2}) = 2^{\frac{1}{2} \cdot 4} - 4 {(\sqrt{2})}^{2}$

Combine $\frac{1}{2}$ and $4$ .

$f (\sqrt{2}) = 2^{\frac{4}{2}} - 4 {(\sqrt{2})}^{2}$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4$ .

$f (\sqrt{2}) = 2^{\frac{2 \cdot 2}{2}} - 4 {(\sqrt{2})}^{2}$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (\sqrt{2}) = 2^{\frac{2 \cdot 2}{2 (1)}} - 4 {(\sqrt{2})}^{2}$

Cancel the common factor.

$f (\sqrt{2}) = 2^{\frac{2 \cdot 2}{2 \cdot 1}} - 4 {(\sqrt{2})}^{2}$

Rewrite the expression.

$f (\sqrt{2}) = 2^{\frac{2}{1}} - 4 {(\sqrt{2})}^{2}$

Divide $2$ by $1$ .

$f (\sqrt{2}) = 2^{2} - 4 {(\sqrt{2})}^{2}$

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

$f (\sqrt{2}) = 4 - 4 {(\sqrt{2})}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$f (\sqrt{2}) = 4 - 4 {(2^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\sqrt{2}) = 4 - 4 \cdot 2^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$f (\sqrt{2}) = 4 - 4 \cdot 2^{\frac{2}{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (\sqrt{2}) = 4 - 4 \cdot 2^{\frac{2}{2}}$

Rewrite the expression.

$f (\sqrt{2}) = 4 - 4 \cdot 2$

Evaluate the exponent.

$f (\sqrt{2}) = 4 - 4 \cdot 2$

Multiply $- 4$ by $2$ .

$f (\sqrt{2}) = 4 - 8$

Subtract $8$ from $4$ .

$f (\sqrt{2}) = - 4$

The final answer is $- 4$ .

$y = - 4$

Step 16

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

$12 {(- \sqrt{2})}^{2} - 8$

Step 17

Evaluate the second derivative.

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Simplify each term.

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Apply the product rule to $- \sqrt{2}$ .

$12 ({(- 1)}^{2} {\sqrt{2}}^{2}) - 8$

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

$12 (1 {\sqrt{2}}^{2}) - 8$

Multiply ${\sqrt{2}}^{2}$ by $1$ .

$12 {\sqrt{2}}^{2} - 8$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$12 {(2^{\frac{1}{2}})}^{2} - 8$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$12 \cdot 2^{\frac{1}{2} \cdot 2} - 8$

Combine $\frac{1}{2}$ and $2$ .

$12 \cdot 2^{\frac{2}{2}} - 8$

Cancel the common factor of $2$ .

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Cancel the common factor.

$12 \cdot 2^{\frac{2}{2}} - 8$

Rewrite the expression.

$12 \cdot 2^{1} - 8$

Evaluate the exponent.

$12 \cdot 2 - 8$

Multiply $12$ by $2$ .

$24 - 8$

Subtract $8$ from $24$ .

$16$

Step 18

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

$x = - \sqrt{2}$ is a local minimum

Step 19

Find the y-value when $x = - \sqrt{2}$ .

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Replace the variable $x$ with $- \sqrt{2}$ in the expression.

$f (- \sqrt{2}) = {(- \sqrt{2})}^{4} - 4 {(- \sqrt{2})}^{2}$

Simplify the result.

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Simplify each term.

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Apply the product rule to $- \sqrt{2}$ .

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

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

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

Multiply ${\sqrt{2}}^{4}$ by $1$ .

$f (- \sqrt{2}) = {\sqrt{2}}^{4} - 4 {(- \sqrt{2})}^{2}$

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

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$f (- \sqrt{2}) = {(2^{\frac{1}{2}})}^{4} - 4 {(- \sqrt{2})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \sqrt{2}) = 2^{\frac{1}{2} \cdot 4} - 4 {(- \sqrt{2})}^{2}$

Combine $\frac{1}{2}$ and $4$ .

$f (- \sqrt{2}) = 2^{\frac{4}{2}} - 4 {(- \sqrt{2})}^{2}$

Cancel the common factor of $4$ and $2$ .

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Factor $2$ out of $4$ .

$f (- \sqrt{2}) = 2^{\frac{2 \cdot 2}{2}} - 4 {(- \sqrt{2})}^{2}$

Cancel the common factors.

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Factor $2$ out of $2$ .

$f (- \sqrt{2}) = 2^{\frac{2 \cdot 2}{2 (1)}} - 4 {(- \sqrt{2})}^{2}$

Cancel the common factor.

$f (- \sqrt{2}) = 2^{\frac{2 \cdot 2}{2 \cdot 1}} - 4 {(- \sqrt{2})}^{2}$

Rewrite the expression.

$f (- \sqrt{2}) = 2^{\frac{2}{1}} - 4 {(- \sqrt{2})}^{2}$

Divide $2$ by $1$ .

$f (- \sqrt{2}) = 2^{2} - 4 {(- \sqrt{2})}^{2}$

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

$f (- \sqrt{2}) = 4 - 4 {(- \sqrt{2})}^{2}$

Apply the product rule to $- \sqrt{2}$ .

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

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

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

Multiply ${\sqrt{2}}^{2}$ by $1$ .

$f (- \sqrt{2}) = 4 - 4 {\sqrt{2}}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$f (- \sqrt{2}) = 4 - 4 {(2^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (- \sqrt{2}) = 4 - 4 \cdot 2^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$f (- \sqrt{2}) = 4 - 4 \cdot 2^{\frac{2}{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$f (- \sqrt{2}) = 4 - 4 \cdot 2^{\frac{2}{2}}$

Rewrite the expression.

$f (- \sqrt{2}) = 4 - 4 \cdot 2$

Evaluate the exponent.

$f (- \sqrt{2}) = 4 - 4 \cdot 2$

Multiply $- 4$ by $2$ .

$f (- \sqrt{2}) = 4 - 8$

Subtract $8$ from $4$ .

$f (- \sqrt{2}) = - 4$

The final answer is $- 4$ .

$y = - 4$

Step 20

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

$(0, 0)$ is a local maxima

$(\sqrt{2}, - 4)$ is a local minima

$(- \sqrt{2}, - 4)$ is a local minima

Step 21