Calculus Examples

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

Find the derivative.

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

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

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

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

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$4 x^{3} + 0$

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

$4 x^{3}$

Set the derivative equal to $0$ .

$4 x^{3} = 0$

Solve for $x$ .

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Divide each term by $4$ and simplify.

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Divide each term in $4 x^{3} = 0$ by $4$ .

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 x^{3}}{4} = \frac{0}{4}$

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{4}$

Divide $0$ by $4$ .

$x^{3} = 0$

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

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

After finding the point that makes the derivative $f' (x) = 4 x^{3}$ equal to $0$ or undefined, the interval to check where $f (x) = x^{4} - 6$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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

$f' (- 1) = 4 {(- 1)}^{3}$

Simplify the result.

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Raise $- 1$ to the power of $3$ .

$f' (- 1) = 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$f' (- 1) = - 4$

The final answer is $- 4$ .

$- 4$

At $x = - 1$ the derivative is $- 4$ . Since this is negative, the function is decreasing on $(- \infty, 0)$ .

Decreasing on $(- \infty, 0)$ since $f' (x) < 0$

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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

$f' (1) = 4 {(1)}^{3}$

Simplify the result.

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One to any power is one.

$f' (1) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f' (1) = 4$

The final answer is $4$ .

$4$

At $x = 1$ the derivative is $4$ . Since this is positive, the function is increasing on $(0, \infty)$ .

Increasing on $(0, \infty)$ since $f' (x) > 0$

List the intervals on which the function is increasing and decreasing.

Increasing on: $(0, \infty)$

Decreasing on: $(- \infty, 0)$

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