Calculus Examples

$y = x^{4} - 2 x^{2}$

Step 1

Write $y = x^{4} - 2 x^{2}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $2$ by $- 2$ .

$4 x^{3} - 4 x$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 4 x)$

Step 3.3

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

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Step 3.3.1

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

$f'' (x) = 12 x^{2} - 4 \frac{d}{d x} x$

Step 3.3.2

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

$f'' (x) = 12 x^{2} - 4 \cdot 1$

Step 3.3.3

Multiply $- 4$ by $1$ .

$f'' (x) = 12 x^{2} - 4$

Step 4

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

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

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.

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Step 5.1.1.1

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

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

Step 5.1.1.2

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

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

Step 5.1.2

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

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Step 5.1.2.1

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

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Multiply $2$ by $- 2$ .

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

Step 5.2

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

$4 x^{3} - 4 x$

Step 6

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

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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

Factor the left side of the equation.

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Step 6.2.1

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

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Step 6.2.1.1

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

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

Step 6.2.1.2

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

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

Step 6.2.1.3

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 1)$ .

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

Step 6.2.2

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

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

Step 6.2.3

Factor.

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Step 6.2.3.1

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 $b = 1$ .

$4 x ((x + 1) (x - 1)) = 0$

Step 6.2.3.2

Remove unnecessary parentheses.

$4 x (x + 1) (x - 1) = 0$

Step 6.3

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

$x = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 6.4

Set $x$ equal to $0$ .

$x = 0$

Step 6.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 6.5.1

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

$x + 1 = 0$

Step 6.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6.6

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 6.6.1

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

$x - 1 = 0$

Step 6.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 6.7

The final solution is all the values that make $4 x (x + 1) (x - 1) = 0$ true.

$x = 0, - 1, 1$

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.

$x = 0, - 1, 1$

Step 9

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.

$12 {(0)}^{2} - 4$

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 $0$ .

$12 \cdot 0 - 4$

Step 10.1.2

Multiply $12$ by $0$ .

$0 - 4$

Step 10.2

Subtract $4$ from $0$ .

$- 4$

Step 11

$x = 0$ 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 12

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

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Step 12.1

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

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

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

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

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

Step 12.2.1.2

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

$f (0) = 0 - 2 \cdot 0$

Step 12.2.1.3

Multiply $- 2$ by $0$ .

$f (0) = 0 + 0$

Step 12.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 12.2.3

The final answer is $0$ .

$y = 0$

Step 13

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.

$12 {(- 1)}^{2} - 4$

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

$12 \cdot 1 - 4$

Step 14.1.2

Multiply $12$ by $1$ .

$12 - 4$

Step 14.2

Subtract $4$ from $12$ .

$8$

Step 15

$x = - 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 1$ is a local minimum

Step 16

Find the y-value when $x = - 1$ .

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Step 16.1

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

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

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

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

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

Step 16.2.1.2

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

$f (- 1) = 1 - 2 \cdot 1$

Step 16.2.1.3

Multiply $- 2$ by $1$ .

$f (- 1) = 1 - 2$

Step 16.2.2

Subtract $2$ from $1$ .

$f (- 1) = - 1$

Step 16.2.3

The final answer is $- 1$ .

$y = - 1$

Step 17

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.

$12 {(1)}^{2} - 4$

Step 18

Evaluate the second derivative.

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Step 18.1

Simplify each term.

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Step 18.1.1

One to any power is one.

$12 \cdot 1 - 4$

Step 18.1.2

Multiply $12$ by $1$ .

$12 - 4$

Step 18.2

Subtract $4$ from $12$ .

$8$

Step 19

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Step 20

Find the y-value when $x = 1$ .

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Step 20.1

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

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

Step 20.2

Simplify the result.

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Step 20.2.1

Simplify each term.

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Step 20.2.1.1

One to any power is one.

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

Step 20.2.1.2

One to any power is one.

$f (1) = 1 - 2 \cdot 1$

Step 20.2.1.3

Multiply $- 2$ by $1$ .

$f (1) = 1 - 2$

Step 20.2.2

Subtract $2$ from $1$ .

$f (1) = - 1$

Step 20.2.3

The final answer is $- 1$ .

$y = - 1$

Step 21

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

$(0, 0)$ is a local maxima

$(- 1, - 1)$ is a local minima

$(1, - 1)$ is a local minima

Step 22