Calculus Examples

$f (x) = 4 {(\frac{x}{x + 5})}^{2}$

Step 1

Find the first derivative of the function.

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Step 1.1

Since $4$ is constant with respect to $x$ , the derivative of $4 {(\frac{x}{x + 5})}^{2}$ with respect to $x$ is $4 \frac{d}{d x} [{(\frac{x}{x + 5})}^{2}]$ .

$4 \frac{d}{d x} [{(\frac{x}{x + 5})}^{2}]$

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \frac{x}{x + 5}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $\frac{x}{x + 5}$ .

$4 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\frac{x}{x + 5}])$

Step 1.2.2

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

$4 (2 u \frac{d}{d x} [\frac{x}{x + 5}])$

Step 1.2.3

Replace all occurrences of $u$ with $\frac{x}{x + 5}$ .

$4 (2 \frac{x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}])$

Step 1.3

Combine fractions.

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Step 1.3.1

Combine $2$ and $\frac{x}{x + 5}$ .

$4 (\frac{2 x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}])$

Step 1.3.2

Combine $\frac{2 x}{x + 5}$ and $4$ .

$\frac{2 x \cdot 4}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}]$

Step 1.3.3

Multiply $4$ by $2$ .

$\frac{8 x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}]$

Step 1.4

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x + 5$ .

$\frac{8 x}{x + 5} \cdot \frac{(x + 5) \frac{d}{d x} [x] - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 1.5

Differentiate.

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Step 1.5.1

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

$\frac{8 x}{x + 5} \cdot \frac{(x + 5) \cdot 1 - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 1.5.2

Multiply $x + 5$ by $1$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 1.5.3

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

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (\frac{d}{d x} [x] + \frac{d}{d x} [5])}{{(x + 5)}^{2}}$

Step 1.5.4

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

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (1 + \frac{d}{d x} [5])}{{(x + 5)}^{2}}$

Step 1.5.5

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (1 + 0)}{{(x + 5)}^{2}}$

Step 1.5.6

Simplify terms.

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Step 1.5.6.1

Add $1$ and $0$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x \cdot 1}{{(x + 5)}^{2}}$

Step 1.5.6.2

Multiply $- 1$ by $1$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x}{{(x + 5)}^{2}}$

Step 1.5.6.3

Subtract $x$ from $x$ .

$\frac{8 x}{x + 5} \cdot \frac{0 + 5}{{(x + 5)}^{2}}$

Step 1.5.6.4

Add $0$ and $5$ .

$\frac{8 x}{x + 5} \cdot \frac{5}{{(x + 5)}^{2}}$

Step 1.5.6.5

Multiply $\frac{8 x}{x + 5}$ by $\frac{5}{{(x + 5)}^{2}}$ .

$\frac{8 x \cdot 5}{(x + 5) {(x + 5)}^{2}}$

Step 1.5.6.6

Multiply $5$ by $8$ .

$\frac{40 x}{(x + 5) {(x + 5)}^{2}}$

Step 1.6

Multiply $x + 5$ by ${(x + 5)}^{2}$ by adding the exponents.

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Step 1.6.1

Multiply $x + 5$ by ${(x + 5)}^{2}$ .

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Step 1.6.1.1

Raise $x + 5$ to the power of $1$ .

$\frac{40 x}{{(x + 5)}^{1} {(x + 5)}^{2}}$

Step 1.6.1.2

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

$\frac{40 x}{{(x + 5)}^{1 + 2}}$

Step 1.6.2

Add $1$ and $2$ .

$\frac{40 x}{{(x + 5)}^{3}}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $40$ is constant with respect to $x$ , the derivative of $\frac{40 x}{{(x + 5)}^{3}}$ with respect to $x$ is $40 \frac{d}{d x} [\frac{x}{{(x + 5)}^{3}}]$ .

$f'' (x) = 40 \frac{d}{d x} (\frac{x}{{(x + 5)}^{3}})$

Step 2.2

$f'' (x) = 40 (\frac{{(x + 5)}^{3} \frac{d}{d x} (x) - x \frac{d}{d x} {(x + 5)}^{3}}{{({(x + 5)}^{3})}^{2}})$

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

Multiply the exponents in ${({(x + 5)}^{3})}^{2}$ .

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Step 2.3.1.1

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

$f'' (x) = 40 (\frac{{(x + 5)}^{3} \frac{d}{d x} (x) - x \frac{d}{d x} {(x + 5)}^{3}}{{(x + 5)}^{3 \cdot 2}})$

Step 2.3.1.2

Multiply $3$ by $2$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{3} \frac{d}{d x} (x) - x \frac{d}{d x} {(x + 5)}^{3}}{{(x + 5)}^{6}})$

Step 2.3.2

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

$f'' (x) = 40 (\frac{{(x + 5)}^{3} \cdot 1 - x \frac{d}{d x} {(x + 5)}^{3}}{{(x + 5)}^{6}})$

Step 2.3.3

Multiply ${(x + 5)}^{3}$ by $1$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{3} - x \frac{d}{d x} {(x + 5)}^{3}}{{(x + 5)}^{6}})$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = x + 5$ .

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Step 2.4.1

To apply the Chain Rule, set $u$ as $x + 5$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{3} - x (\frac{d}{d u} (u^{3}) \frac{d}{d x} (x + 5))}{{(x + 5)}^{6}})$

Step 2.4.2

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

$f'' (x) = 40 (\frac{{(x + 5)}^{3} - x (3 u^{2} \frac{d}{d x} (x + 5))}{{(x + 5)}^{6}})$

Step 2.4.3

Replace all occurrences of $u$ with $x + 5$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{3} - x (3 {(x + 5)}^{2} \frac{d}{d x} (x + 5))}{{(x + 5)}^{6}})$

Step 2.5

Simplify with factoring out.

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Step 2.5.1

Multiply $3$ by $- 1$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{3} - 3 x ({(x + 5)}^{2} \frac{d}{d x} (x + 5))}{{(x + 5)}^{6}})$

Step 2.5.2

Factor ${(x + 5)}^{2}$ out of ${(x + 5)}^{3} - 3 x ({(x + 5)}^{2} \frac{d}{d x} [x + 5])$ .

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Step 2.5.2.1

Factor ${(x + 5)}^{2}$ out of ${(x + 5)}^{3}$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{2} (x + 5) - 3 x ({(x + 5)}^{2} \frac{d}{d x} (x + 5))}{{(x + 5)}^{6}})$

Step 2.5.2.2

Factor ${(x + 5)}^{2}$ out of $- 3 x ({(x + 5)}^{2} \frac{d}{d x} [x + 5])$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{2} (x + 5) + {(x + 5)}^{2} (- 3 x (\frac{d}{d x} (x + 5)))}{{(x + 5)}^{6}})$

Step 2.5.2.3

Factor ${(x + 5)}^{2}$ out of ${(x + 5)}^{2} (x + 5) + {(x + 5)}^{2} (- 3 x (\frac{d}{d x} [x + 5]))$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{2} (x + 5 - 3 x (\frac{d}{d x} (x + 5)))}{{(x + 5)}^{6}})$

Step 2.6

Cancel the common factors.

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Step 2.6.1

Factor ${(x + 5)}^{2}$ out of ${(x + 5)}^{6}$ .

$f'' (x) = 40 (\frac{{(x + 5)}^{2} (x + 5 - 3 x (\frac{d}{d x} (x + 5)))}{{(x + 5)}^{2} {(x + 5)}^{4}})$

Step 2.6.2

Cancel the common factor.

$f'' (x) = 40 (\frac{{(x + 5)}^{2} (x + 5 - 3 x (\frac{d}{d x} (x + 5)))}{{(x + 5)}^{2} {(x + 5)}^{4}})$

Step 2.6.3

Rewrite the expression.

$f'' (x) = 40 (\frac{x + 5 - 3 x (\frac{d}{d x} (x + 5))}{{(x + 5)}^{4}})$

Step 2.7

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

$f'' (x) = 40 (\frac{x + 5 - 3 x (\frac{d}{d x} (x) + \frac{d}{d x} (5))}{{(x + 5)}^{4}})$

Step 2.8

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

$f'' (x) = 40 (\frac{x + 5 - 3 x (1 + \frac{d}{d x} (5))}{{(x + 5)}^{4}})$

Step 2.9

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$f'' (x) = 40 (\frac{x + 5 - 3 x (1 + 0)}{{(x + 5)}^{4}})$

Step 2.10

Simplify terms.

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Step 2.10.1

Add $1$ and $0$ .

$f'' (x) = 40 (\frac{x + 5 - 3 x \cdot 1}{{(x + 5)}^{4}})$

Step 2.10.2

Multiply $- 3$ by $1$ .

$f'' (x) = 40 (\frac{x + 5 - 3 x}{{(x + 5)}^{4}})$

Step 2.10.3

Subtract $3 x$ from $x$ .

$f'' (x) = 40 (\frac{- 2 x + 5}{{(x + 5)}^{4}})$

Step 2.10.4

Combine $40$ and $\frac{- 2 x + 5}{{(x + 5)}^{4}}$ .

$f'' (x) = \frac{40 (- 2 x + 5)}{{(x + 5)}^{4}}$

Step 2.11

Simplify.

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Step 2.11.1

Apply the distributive property.

$f'' (x) = \frac{40 (- 2 x) + 40 \cdot 5}{{(x + 5)}^{4}}$

Step 2.11.2

Simplify each term.

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Step 2.11.2.1

Multiply $- 2$ by $40$ .

$f'' (x) = \frac{- 80 x + 40 \cdot 5}{{(x + 5)}^{4}}$

Step 2.11.2.2

Multiply $40$ by $5$ .

$f'' (x) = \frac{- 80 x + 200}{{(x + 5)}^{4}}$

Step 2.11.3

Factor $40$ out of $- 80 x + 200$ .

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Step 2.11.3.1

Factor $40$ out of $- 80 x$ .

$f'' (x) = \frac{40 (- 2 x) + 200}{{(x + 5)}^{4}}$

Step 2.11.3.2

Factor $40$ out of $200$ .

$f'' (x) = \frac{40 (- 2 x) + 40 (5)}{{(x + 5)}^{4}}$

Step 2.11.3.3

Factor $40$ out of $40 (- 2 x) + 40 (5)$ .

$f'' (x) = \frac{40 (- 2 x + 5)}{{(x + 5)}^{4}}$

Step 2.11.4

Factor $- 1$ out of $- 2 x$ .

$f'' (x) = \frac{40 (- (2 x) + 5)}{{(x + 5)}^{4}}$

Step 2.11.5

Rewrite $5$ as $- 1 (- 5)$ .

$f'' (x) = \frac{40 (- (2 x) - 1 \cdot - 5)}{{(x + 5)}^{4}}$

Step 2.11.6

Factor $- 1$ out of $- (2 x) - 1 (- 5)$ .

$f'' (x) = \frac{40 (- (2 x - 5))}{{(x + 5)}^{4}}$

Step 2.11.7

Rewrite $- (2 x - 5)$ as $- 1 (2 x - 5)$ .

$f'' (x) = \frac{40 (- 1 (2 x - 5))}{{(x + 5)}^{4}}$

Step 2.11.8

Move the negative in front of the fraction.

$f'' (x) = - \frac{40 (2 x - 5)}{{(x + 5)}^{4}}$

Step 3

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

$\frac{40 x}{{(x + 5)}^{3}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Since $4$ is constant with respect to $x$ , the derivative of $4 {(\frac{x}{x + 5})}^{2}$ with respect to $x$ is $4 \frac{d}{d x} [{(\frac{x}{x + 5})}^{2}]$ .

$4 \frac{d}{d x} [{(\frac{x}{x + 5})}^{2}]$

Step 4.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \frac{x}{x + 5}$ .

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Step 4.1.2.1

To apply the Chain Rule, set $u$ as $\frac{x}{x + 5}$ .

$4 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\frac{x}{x + 5}])$

Step 4.1.2.2

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

$4 (2 u \frac{d}{d x} [\frac{x}{x + 5}])$

Step 4.1.2.3

Replace all occurrences of $u$ with $\frac{x}{x + 5}$ .

$4 (2 \frac{x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}])$

Step 4.1.3

Combine fractions.

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Step 4.1.3.1

Combine $2$ and $\frac{x}{x + 5}$ .

$4 (\frac{2 x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}])$

Step 4.1.3.2

Combine $\frac{2 x}{x + 5}$ and $4$ .

$\frac{2 x \cdot 4}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}]$

Step 4.1.3.3

Multiply $4$ by $2$ .

$\frac{8 x}{x + 5} \frac{d}{d x} [\frac{x}{x + 5}]$

Step 4.1.4

$\frac{8 x}{x + 5} \cdot \frac{(x + 5) \frac{d}{d x} [x] - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 4.1.5

Differentiate.

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Step 4.1.5.1

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

$\frac{8 x}{x + 5} \cdot \frac{(x + 5) \cdot 1 - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 4.1.5.2

Multiply $x + 5$ by $1$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x \frac{d}{d x} [x + 5]}{{(x + 5)}^{2}}$

Step 4.1.5.3

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

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (\frac{d}{d x} [x] + \frac{d}{d x} [5])}{{(x + 5)}^{2}}$

Step 4.1.5.4

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

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (1 + \frac{d}{d x} [5])}{{(x + 5)}^{2}}$

Step 4.1.5.5

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x (1 + 0)}{{(x + 5)}^{2}}$

Step 4.1.5.6

Simplify terms.

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Step 4.1.5.6.1

Add $1$ and $0$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x \cdot 1}{{(x + 5)}^{2}}$

Step 4.1.5.6.2

Multiply $- 1$ by $1$ .

$\frac{8 x}{x + 5} \cdot \frac{x + 5 - x}{{(x + 5)}^{2}}$

Step 4.1.5.6.3

Subtract $x$ from $x$ .

$\frac{8 x}{x + 5} \cdot \frac{0 + 5}{{(x + 5)}^{2}}$

Step 4.1.5.6.4

Add $0$ and $5$ .

$\frac{8 x}{x + 5} \cdot \frac{5}{{(x + 5)}^{2}}$

Step 4.1.5.6.5

Multiply $\frac{8 x}{x + 5}$ by $\frac{5}{{(x + 5)}^{2}}$ .

$\frac{8 x \cdot 5}{(x + 5) {(x + 5)}^{2}}$

Step 4.1.5.6.6

Multiply $5$ by $8$ .

$\frac{40 x}{(x + 5) {(x + 5)}^{2}}$

Step 4.1.6

Multiply $x + 5$ by ${(x + 5)}^{2}$ by adding the exponents.

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Step 4.1.6.1

Multiply $x + 5$ by ${(x + 5)}^{2}$ .

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Step 4.1.6.1.1

Raise $x + 5$ to the power of $1$ .

$\frac{40 x}{{(x + 5)}^{1} {(x + 5)}^{2}}$

Step 4.1.6.1.2

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

$\frac{40 x}{{(x + 5)}^{1 + 2}}$

Step 4.1.6.2

Add $1$ and $2$ .

$f' (x) = \frac{40 x}{{(x + 5)}^{3}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{40 x}{{(x + 5)}^{3}}$ .

$\frac{40 x}{{(x + 5)}^{3}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{40 x}{{(x + 5)}^{3}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{40 x}{{(x + 5)}^{3}} = 0$

Step 5.2

Set the numerator equal to zero.

$40 x = 0$

Step 5.3

Divide each term in $40 x = 0$ by $40$ and simplify.

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Step 5.3.1

Divide each term in $40 x = 0$ by $40$ .

$\frac{40 x}{40} = \frac{0}{40}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $40$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{40 x}{40} = \frac{0}{40}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{40}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $0$ by $40$ .

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{40 x}{{(x + 5)}^{3}}$ equal to $0$ to find where the expression is undefined.

${(x + 5)}^{3} = 0$

Step 6.2

Solve for $x$ .

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Step 6.2.1

Set the $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 6.2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 7

Critical points to evaluate.

$x = 0$

Step 8

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.

$- \frac{40 (2 (0) - 5)}{{((0) + 5)}^{4}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Multiply $2$ by $0$ .

$- \frac{40 (0 - 5)}{{(0 + 5)}^{4}}$

Step 9.1.2

Subtract $5$ from $0$ .

$- \frac{40 \cdot - 5}{{(0 + 5)}^{4}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Add $0$ and $5$ .

$- \frac{40 \cdot - 5}{5^{4}}$

Step 9.2.2

Raise $5$ to the power of $4$ .

$- \frac{40 \cdot - 5}{625}$

Step 9.3

Reduce the expression by cancelling the common factors.

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Step 9.3.1

Multiply $40$ by $- 5$ .

$- \frac{- 200}{625}$

Step 9.3.2

Cancel the common factor of $- 200$ and $625$ .

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Step 9.3.2.1

Factor $25$ out of $- 200$ .

$- \frac{25 (- 8)}{625}$

Step 9.3.2.2

Cancel the common factors.

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Step 9.3.2.2.1

Factor $25$ out of $625$ .

$- \frac{25 \cdot - 8}{25 \cdot 25}$

Step 9.3.2.2.2

Cancel the common factor.

$- \frac{25 \cdot - 8}{25 \cdot 25}$

Step 9.3.2.2.3

Rewrite the expression.

$- \frac{- 8}{25}$

Step 9.3.3

Move the negative in front of the fraction.

$- - \frac{8}{25}$

Step 9.4

Multiply $- - \frac{8}{25}$ .

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Step 9.4.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{8}{25})$

Step 9.4.2

Multiply $\frac{8}{25}$ by $1$ .

$\frac{8}{25}$

Step 10

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 11

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

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Step 11.1

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

$f (0) = 4 {(\frac{0}{(0) + 5})}^{2}$

Step 11.2

Simplify the result.

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Step 11.2.1

Cancel the common factor of $0$ and $(0) + 5$ .

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Step 11.2.1.1

Factor $5$ out of $0$ .

$f (0) = 4 {(\frac{5 (0)}{(0) + 5})}^{2}$

Step 11.2.1.2

Cancel the common factors.

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Step 11.2.1.2.1

Factor $5$ out of $0$ .

$f (0) = 4 {(\frac{5 \cdot 0}{5 \cdot 0 + 5})}^{2}$

Step 11.2.1.2.2

Factor $5$ out of $5$ .

$f (0) = 4 {(\frac{5 (0)}{5 \cdot 0 + 5 \cdot 1})}^{2}$

Step 11.2.1.2.3

Factor $5$ out of $5 \cdot 0 + 5 \cdot 1$ .

$f (0) = 4 {(\frac{5 (0)}{5 \cdot (0 + 1)})}^{2}$

Step 11.2.1.2.4

Cancel the common factor.

$f (0) = 4 {(\frac{5 \cdot 0}{5 \cdot (0 + 1)})}^{2}$

Step 11.2.1.2.5

Rewrite the expression.

$f (0) = 4 {(\frac{0}{0 + 1})}^{2}$

Step 11.2.2

Simplify the expression.

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Step 11.2.2.1

Add $0$ and $1$ .

$f (0) = 4 {(\frac{0}{1})}^{2}$

Step 11.2.2.2

Divide $0$ by $1$ .

$f (0) = 4 \cdot 0^{2}$

Step 11.2.2.3

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

$f (0) = 4 \cdot 0$

Step 11.2.2.4

Multiply $4$ by $0$ .

$f (0) = 0$

Step 11.2.3

The final answer is $0$ .

$y = 0$

Step 12

These are the local extrema for $f (x) = 4 {(\frac{x}{x + 5})}^{2}$ .

$(0, 0)$ is a local minima

Step 13