Calculus Examples

$f (x) = x^{4} - 6 x^{2}$

Find the first derivative of the function.

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By the Sum Rule, the derivative of $x^{4} - 6 x^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 6 x^{2}]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 6 x^{2}]$

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

$4 x^{3} + \frac{d}{d x} [- 6 x^{2}]$

Evaluate $\frac{d}{d x} [- 6 x^{2}]$ .

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Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x^{2}$ with respect to $x$ is $- 6 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 6 \frac{d}{d x} [x^{2}]$

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

$4 x^{3} - 6 (2 x)$

Multiply $2$ by $- 6$ .

$4 x^{3} - 12 x$

Find the second derivative of the function.

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By the Sum Rule, the derivative of $4 x^{3} - 12 x$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 12 x]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 12 x)$

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

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 12 x)$

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

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 12 x)$

Multiply $3$ by $4$ .

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

Evaluate $\frac{d}{d x} [- 12 x]$ .

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Since $- 12$ is constant with respect to $x$ , the derivative of $- 12 x$ with respect to $x$ is $- 12 \frac{d}{d x} [x]$ .

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

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

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

Multiply $- 12$ by $1$ .

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

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

$4 x^{3} - 12 x = 0$

Factor $4 x$ out of $4 x^{3} - 12 x$ .

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

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

Factor $4 x$ out of $- 12 x$ .

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

Factor $4 x$ out of $4 x (x^{2}) + 4 x (- 3)$ .

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

Divide each term by $4$ and simplify.

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

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

Reduce the expression by cancelling the common factors.

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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 $x^{2} - 3$ equal to $0$ and solve for $x$ .

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Set the factor equal to $0$ .

$x^{2} - 3 = 0$

Add $3$ to both sides of the equation.

$x^{2} = 3$

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

The solution is the result of $4 x = 0$ and $x^{2} - 3 = 0$ .

$x = 0, \sqrt{3}, - \sqrt{3}$

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.

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

Evaluate the second derivative.

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Simplify each term.

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Remove parentheses around $0$ .

$12 \cdot 0^{2} - 12$

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

$12 \cdot 0 - 12$

Multiply $12$ by $0$ .

$0 - 12$

Subtract $12$ from $0$ .

$- 12$

$x = 0$ 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

Evaluate the second derivative at $x = \sqrt{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(\sqrt{3})}^{2} - 12$

Evaluate the second derivative.

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Simplify each term.

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Remove parentheses around $\sqrt{3}$ .

$12 {\sqrt{3}}^{2} - 12$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

$12 \cdot 3 - 12$

Multiply $12$ by $3$ .

$36 - 12$

Subtract $12$ from $36$ .

$24$

$x = \sqrt{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \sqrt{3}$ is a local minimum

Evaluate the second derivative at $x = - \sqrt{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(- \sqrt{3})}^{2} - 12$

Evaluate the second derivative.

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Simplify each term.

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Apply the product rule to $- \sqrt{3}$ .

$12 ({(- 1)}^{2} {\sqrt{3}}^{2}) - 12$

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

$12 (1 {\sqrt{3}}^{2}) - 12$

Multiply ${\sqrt{3}}^{2}$ by $1$ .

$12 {\sqrt{3}}^{2} - 12$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

$12 \cdot 3 - 12$

Multiply $12$ by $3$ .

$36 - 12$

Subtract $12$ from $36$ .

$24$

$x = - \sqrt{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - \sqrt{3}$ is a local minimum

These are the local extrema for $f (x) = x^{4} - 6 x^{2}$ .

$x = 0$ is a local maximum

$x = \sqrt{3}$ is a local minimum

$x = - \sqrt{3}$ is a local minimum

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