Calculus Examples

$f (x) = x^{5} - 4$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $x^{5} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 4]$

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$5 x^{4} + \frac{d}{d x} [- 4]$

Step 1.1.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$5 x^{4} + 0$

Step 1.1.4

Add $5 x^{4}$ and $0$ .

$f' (x) = 5 x^{4}$

Step 1.2

Find the second derivative.

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Step 1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{4}$ with respect to $x$ is $5 \frac{d}{d x} [x^{4}]$ .

$5 \frac{d}{d x} [x^{4}]$

Step 1.2.2

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

$5 (4 x^{3})$

Step 1.2.3

Multiply $4$ by $5$ .

$f'' (x) = 20 x^{3}$

Step 1.3

The second derivative of $f (x)$ with respect to $x$ is $20 x^{3}$ .

$20 x^{3}$

Step 2

Set the second derivative equal to $0$ then solve the equation $20 x^{3} = 0$ .

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Step 2.1

Set the second derivative equal to $0$ .

$20 x^{3} = 0$

Step 2.2

Divide each term in $20 x^{3} = 0$ by $20$ and simplify.

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Step 2.2.1

Divide each term in $20 x^{3} = 0$ by $20$ .

$\frac{20 x^{3}}{20} = \frac{0}{20}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $20$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{20 x^{3}}{20} = \frac{0}{20}$

Step 2.2.2.1.2

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

$x^{3} = \frac{0}{20}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $0$ by $20$ .

$x^{3} = 0$

Step 2.3

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

$x = 0$

Step 3

Find the points where the second derivative is $0$ .

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Step 3.1

Substitute $0$ in $f (x) = x^{5} - 4$ to find the value of $y$ .

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Step 3.1.1

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

$f (0) = {(0)}^{5} - 4$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

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

$f (0) = 0 - 4$

Step 3.1.2.2

Subtract $4$ from $0$ .

$f (0) = - 4$

Step 3.1.2.3

The final answer is $- 4$ .

$- 4$

Step 3.2

The point found by substituting $0$ in $f (x) = x^{5} - 4$ is $(0, - 4)$ . This point can be an inflection point.

$(0, - 4)$

Step 4

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

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

Step 5

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

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Step 5.1

Replace the variable $x$ with $- 0.1$ in the expression.

$f'' (- 0.1) = 20 {(- 0.1)}^{3}$

Step 5.2

Simplify the result.

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Step 5.2.1

Raise $- 0.1$ to the power of $3$ .

$f'' (- 0.1) = 20 \cdot - 0.001$

Step 5.2.2

Multiply $20$ by $- 0.001$ .

$f'' (- 0.1) = - 0.02$

Step 5.2.3

The final answer is $- 0.02$ .

$- 0.02$

Step 5.3

At $- 0.1$ , the second derivative is $- 0.02$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 0)$

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

Step 6

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

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Step 6.1

Replace the variable $x$ with $0.1$ in the expression.

$f'' (0.1) = 20 {(0.1)}^{3}$

Step 6.2

Simplify the result.

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Step 6.2.1

Raise $0.1$ to the power of $3$ .

$f'' (0.1) = 20 \cdot 0.001$

Step 6.2.2

Multiply $20$ by $0.001$ .

$f'' (0.1) = 0.02$

Step 6.2.3

The final answer is $0.02$ .

$0.02$

Step 6.3

At $0.1$ , the second derivative is $0.02$ . Since this is positive, the second derivative is increasing on the interval $(0, \infty)$ .

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

Step 7

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(0, - 4)$ .

$(0, - 4)$

Step 8

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