Calculus Examples

$f (x) = x^{2} + \frac{320}{x}$ ; $(0, \infty)$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate.

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Step 1.1.1.1.1

By the Sum Rule, the derivative of $x^{2} + \frac{320}{x}$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [\frac{320}{x}]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [\frac{320}{x}]$

Step 1.1.1.1.2

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

$2 x + \frac{d}{d x} [\frac{320}{x}]$

Step 1.1.1.2

Evaluate $\frac{d}{d x} [\frac{320}{x}]$ .

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Step 1.1.1.2.1

Since $320$ is constant with respect to $x$ , the derivative of $\frac{320}{x}$ with respect to $x$ is $320 \frac{d}{d x} [\frac{1}{x}]$ .

$2 x + 320 \frac{d}{d x} [\frac{1}{x}]$

Step 1.1.1.2.2

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$2 x + 320 \frac{d}{d x} [x^{- 1}]$

Step 1.1.1.2.3

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

$2 x + 320 (- x^{- 2})$

Step 1.1.1.2.4

Multiply $- 1$ by $320$ .

$2 x - 320 x^{- 2}$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2 x - 320 \frac{1}{x^{2}}$

Step 1.1.1.3.2

Combine terms.

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Step 1.1.1.3.2.1

Combine $- 320$ and $\frac{1}{x^{2}}$ .

$2 x + \frac{- 320}{x^{2}}$

Step 1.1.1.3.2.2

Move the negative in front of the fraction.

$f' (x) = 2 x - \frac{320}{x^{2}}$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $2 x - \frac{320}{x^{2}}$ .

$2 x - \frac{320}{x^{2}}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $2 x - \frac{320}{x^{2}} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$2 x - \frac{320}{x^{2}} = 0$

Step 1.2.2

Find the LCD of the terms in the equation.

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Step 1.2.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x^{2}, 1$

Step 1.2.2.2

The LCM of one and any expression is the expression.

$x^{2}$

Step 1.2.3

Multiply each term in $2 x - \frac{320}{x^{2}} = 0$ by $x^{2}$ to eliminate the fractions.

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Step 1.2.3.1

Multiply each term in $2 x - \frac{320}{x^{2}} = 0$ by $x^{2}$ .

$2 x \cdot x^{2} - \frac{320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Simplify each term.

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Step 1.2.3.2.1.1

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 1.2.3.2.1.1.1

Move $x^{2}$ .

$2 (x^{2} x) - \frac{320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.1.2

Multiply $x^{2}$ by $x$ .

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Step 1.2.3.2.1.1.2.1

Raise $x$ to the power of $1$ .

$2 (x^{2} x^{1}) - \frac{320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 x^{2 + 1} - \frac{320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.1.3

Add $2$ and $1$ .

$2 x^{3} - \frac{320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.2

Cancel the common factor of $x^{2}$ .

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Step 1.2.3.2.1.2.1

Move the leading negative in $- \frac{320}{x^{2}}$ into the numerator.

$2 x^{3} + \frac{- 320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$2 x^{3} + \frac{- 320}{x^{2}} x^{2} = 0 x^{2}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$2 x^{3} - 320 = 0 x^{2}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Multiply $0$ by $x^{2}$ .

$2 x^{3} - 320 = 0$

Step 1.2.4

Solve the equation.

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Step 1.2.4.1

Add $320$ to both sides of the equation.

$2 x^{3} = 320$

Step 1.2.4.2

Subtract $320$ from both sides of the equation.

$2 x^{3} - 320 = 0$

Step 1.2.4.3

Factor $2$ out of $2 x^{3} - 320$ .

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Step 1.2.4.3.1

Factor $2$ out of $2 x^{3}$ .

$2 (x^{3}) - 320 = 0$

Step 1.2.4.3.2

Factor $2$ out of $- 320$ .

$2 x^{3} + 2 \cdot - 160 = 0$

Step 1.2.4.3.3

Factor $2$ out of $2 x^{3} + 2 \cdot - 160$ .

$2 (x^{3} - 160) = 0$

Step 1.2.4.4

Divide each term in $2 (x^{3} - 160) = 0$ by $2$ and simplify.

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Step 1.2.4.4.1

Divide each term in $2 (x^{3} - 160) = 0$ by $2$ .

$\frac{2 (x^{3} - 160)}{2} = \frac{0}{2}$

Step 1.2.4.4.2

Simplify the left side.

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Step 1.2.4.4.2.1

Cancel the common factor of $2$ .

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Step 1.2.4.4.2.1.1

Cancel the common factor.

$\frac{2 (x^{3} - 160)}{2} = \frac{0}{2}$

Step 1.2.4.4.2.1.2

Divide $x^{3} - 160$ by $1$ .

$x^{3} - 160 = \frac{0}{2}$

Step 1.2.4.4.3

Simplify the right side.

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Step 1.2.4.4.3.1

Divide $0$ by $2$ .

$x^{3} - 160 = 0$

Step 1.2.4.5

Add $160$ to both sides of the equation.

$x^{3} = 160$

Step 1.2.4.6

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

$x = \sqrt[3]{160}$

Step 1.2.4.7

Simplify $\sqrt[3]{160}$ .

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Step 1.2.4.7.1

Rewrite $160$ as $2^{3} \cdot 20$ .

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Step 1.2.4.7.1.1

Factor $8$ out of $160$ .

$x = \sqrt[3]{8 (20)}$

Step 1.2.4.7.1.2

Rewrite $8$ as $2^{3}$ .

$x = \sqrt[3]{2^{3} \cdot 20}$

Step 1.2.4.7.2

Pull terms out from under the radical.

$x = 2 \sqrt[3]{20}$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the denominator in $\frac{320}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Step 1.3.2

Solve for $x$ .

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Step 1.3.2.1

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

$x = \pm \sqrt{0}$

Step 1.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 1.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.3.2.2.2

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

$x = \pm 0$

Step 1.3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.4

Evaluate $x^{2} + \frac{320}{x}$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 2 \sqrt[3]{20}$ .

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Step 1.4.1.1

Substitute $2 \sqrt[3]{20}$ for $x$ .

${(2 \sqrt[3]{20})}^{2} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

Apply the product rule to $2 \sqrt[3]{20}$ .

$2^{2} {\sqrt[3]{20}}^{2} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.2

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

$4 {\sqrt[3]{20}}^{2} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.3

Rewrite ${\sqrt[3]{20}}^{2}$ as $\sqrt[3]{20^{2}}$ .

$4 \sqrt[3]{20^{2}} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.4

Raise $20$ to the power of $2$ .

$4 \sqrt[3]{400} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.5

Rewrite $400$ as $2^{3} \cdot 50$ .

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Step 1.4.1.2.1.5.1

Factor $8$ out of $400$ .

$4 \sqrt[3]{8 (50)} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.5.2

Rewrite $8$ as $2^{3}$ .

$4 \sqrt[3]{2^{3} \cdot 50} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.6

Pull terms out from under the radical.

$4 (2 \sqrt[3]{50}) + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.7

Multiply $2$ by $4$ .

$8 \sqrt[3]{50} + \frac{320}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.8

Cancel the common factor of $320$ and $2$ .

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Step 1.4.1.2.1.8.1

Factor $2$ out of $320$ .

$8 \sqrt[3]{50} + \frac{2 \cdot 160}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.8.2

Cancel the common factors.

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Step 1.4.1.2.1.8.2.1

Factor $2$ out of $2 \sqrt[3]{20}$ .

$8 \sqrt[3]{50} + \frac{2 \cdot 160}{2 (\sqrt[3]{20})}$

Step 1.4.1.2.1.8.2.2

Cancel the common factor.

$8 \sqrt[3]{50} + \frac{2 \cdot 160}{2 \sqrt[3]{20}}$

Step 1.4.1.2.1.8.2.3

Rewrite the expression.

$8 \sqrt[3]{50} + \frac{160}{\sqrt[3]{20}}$

Step 1.4.1.2.1.9

Multiply $\frac{160}{\sqrt[3]{20}}$ by $\frac{{\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{2}}$ .

$8 \sqrt[3]{50} + \frac{160}{\sqrt[3]{20}} \cdot \frac{{\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{2}}$

Step 1.4.1.2.1.10

Combine and simplify the denominator.

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Step 1.4.1.2.1.10.1

Multiply $\frac{160}{\sqrt[3]{20}}$ by $\frac{{\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{2}}$ .

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{\sqrt[3]{20} {\sqrt[3]{20}}^{2}}$

Step 1.4.1.2.1.10.2

Raise $\sqrt[3]{20}$ to the power of $1$ .

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{1} {\sqrt[3]{20}}^{2}}$

Step 1.4.1.2.1.10.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{1 + 2}}$

Step 1.4.1.2.1.10.4

Add $1$ and $2$ .

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{{\sqrt[3]{20}}^{3}}$

Step 1.4.1.2.1.10.5

Rewrite ${\sqrt[3]{20}}^{3}$ as $20$ .

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Step 1.4.1.2.1.10.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{20}$ as $20^{\frac{1}{3}}$ .

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{{(20^{\frac{1}{3}})}^{3}}$

Step 1.4.1.2.1.10.5.2

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

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{20^{\frac{1}{3} \cdot 3}}$

Step 1.4.1.2.1.10.5.3

Combine $\frac{1}{3}$ and $3$ .

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{20^{\frac{3}{3}}}$

Step 1.4.1.2.1.10.5.4

Cancel the common factor of $3$ .

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Step 1.4.1.2.1.10.5.4.1

Cancel the common factor.

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{20^{\frac{3}{3}}}$

Step 1.4.1.2.1.10.5.4.2

Rewrite the expression.

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{20^{1}}$

Step 1.4.1.2.1.10.5.5

Evaluate the exponent.

$8 \sqrt[3]{50} + \frac{160 {\sqrt[3]{20}}^{2}}{20}$

Step 1.4.1.2.1.11

Cancel the common factor of $160$ and $20$ .

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Step 1.4.1.2.1.11.1

Factor $20$ out of $160 {\sqrt[3]{20}}^{2}$ .

$8 \sqrt[3]{50} + \frac{20 (8 {\sqrt[3]{20}}^{2})}{20}$

Step 1.4.1.2.1.11.2

Cancel the common factors.

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Step 1.4.1.2.1.11.2.1

Factor $20$ out of $20$ .

$8 \sqrt[3]{50} + \frac{20 (8 {\sqrt[3]{20}}^{2})}{20 (1)}$

Step 1.4.1.2.1.11.2.2

Cancel the common factor.

$8 \sqrt[3]{50} + \frac{20 (8 {\sqrt[3]{20}}^{2})}{20 \cdot 1}$

Step 1.4.1.2.1.11.2.3

Rewrite the expression.

$8 \sqrt[3]{50} + \frac{8 {\sqrt[3]{20}}^{2}}{1}$

Step 1.4.1.2.1.11.2.4

Divide $8 {\sqrt[3]{20}}^{2}$ by $1$ .

$8 \sqrt[3]{50} + 8 {\sqrt[3]{20}}^{2}$

Step 1.4.1.2.1.12

Rewrite ${\sqrt[3]{20}}^{2}$ as $\sqrt[3]{20^{2}}$ .

$8 \sqrt[3]{50} + 8 \sqrt[3]{20^{2}}$

Step 1.4.1.2.1.13

Raise $20$ to the power of $2$ .

$8 \sqrt[3]{50} + 8 \sqrt[3]{400}$

Step 1.4.1.2.1.14

Rewrite $400$ as $2^{3} \cdot 50$ .

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Step 1.4.1.2.1.14.1

Factor $8$ out of $400$ .

$8 \sqrt[3]{50} + 8 \sqrt[3]{8 (50)}$

Step 1.4.1.2.1.14.2

Rewrite $8$ as $2^{3}$ .

$8 \sqrt[3]{50} + 8 \sqrt[3]{2^{3} \cdot 50}$

Step 1.4.1.2.1.15

Pull terms out from under the radical.

$8 \sqrt[3]{50} + 8 (2 \sqrt[3]{50})$

Step 1.4.1.2.1.16

Multiply $2$ by $8$ .

$8 \sqrt[3]{50} + 16 \sqrt[3]{50}$

Step 1.4.1.2.2

Add $8 \sqrt[3]{50}$ and $16 \sqrt[3]{50}$ .

$24 \sqrt[3]{50}$

Step 1.4.2

Evaluate at $x = 0$ .

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Step 1.4.2.1

Substitute $0$ for $x$ .

${(0)}^{2} + \frac{320}{0}$

Step 1.4.2.2

The expression contains a division by $0$ . The expression is undefined.

Undefined

Step 1.4.3

List all of the points.

$(2 \sqrt[3]{20}, 24 \sqrt[3]{50})$

Step 2

Use the first derivative test to determine which points can be maxima or minima.

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Step 2.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, 2 \sqrt[3]{20}) \cup (2 \sqrt[3]{20}, \infty)$

Step 2.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 2 \sqrt[3]{20})$ in the first derivative $2 x - \frac{320}{x^{2}}$ to check if the result is negative or positive.

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Step 2.2.1

Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = 2 (- 2) - \frac{320}{{(- 2)}^{2}}$

Step 2.2.2

Simplify the result.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

Multiply $2$ by $- 2$ .

$f' (- 2) = - 4 - \frac{320}{{(- 2)}^{2}}$

Step 2.2.2.1.2

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

$f' (- 2) = - 4 - \frac{320}{4}$

Step 2.2.2.1.3

Divide $320$ by $4$ .

$f' (- 2) = - 4 - 1 \cdot 80$

Step 2.2.2.1.4

Multiply $- 1$ by $80$ .

$f' (- 2) = - 4 - 80$

Step 2.2.2.2

Subtract $80$ from $- 4$ .

$f' (- 2) = - 84$

Step 2.2.2.3

The final answer is $- 84$ .

$- 84$

Step 2.3

Substitute any number, such as $8$ , from the interval $(2 \sqrt[3]{20}, \infty)$ in the first derivative $2 x - \frac{320}{x^{2}}$ to check if the result is negative or positive.

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Step 2.3.1

Replace the variable $x$ with $8$ in the expression.

$f' (8) = 2 (8) - \frac{320}{{(8)}^{2}}$

Step 2.3.2

Simplify the result.

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Step 2.3.2.1

Simplify each term.

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Step 2.3.2.1.1

Multiply $2$ by $8$ .

$f' (8) = 16 - \frac{320}{{(8)}^{2}}$

Step 2.3.2.1.2

Raise $8$ to the power of $2$ .

$f' (8) = 16 - \frac{320}{64}$

Step 2.3.2.1.3

Divide $320$ by $64$ .

$f' (8) = 16 - 1 \cdot 5$

Step 2.3.2.1.4

Multiply $- 1$ by $5$ .

$f' (8) = 16 - 5$

Step 2.3.2.2

Subtract $5$ from $16$ .

$f' (8) = 11$

Step 2.3.2.3

The final answer is $11$ .

$11$

Step 2.4

Since the first derivative changed signs from negative to positive around $x = 2 \sqrt[3]{20}$ , then $x = 2 \sqrt[3]{20}$ is a local minimum.

$x = 2 \sqrt[3]{20}$ is a local minimum

Step 3

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

No absolute maximum

Absolute Minimum: $(2 \sqrt[3]{20}, 24 \sqrt[3]{50})$

Step 4