Calculus Examples

$x {(\frac{x}{2} - 5)}^{4}$

Step 1

Write $x {(\frac{x}{2} - 5)}^{4}$ as a function.

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = {(\frac{x}{2} - 5)}^{4}$ .

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

Step 2.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = \frac{x}{2} - 5$ .

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Step 2.1.2.1

To apply the Chain Rule, set $u$ as $\frac{x}{2} - 5$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Replace all occurrences of $u$ with $\frac{x}{2} - 5$ .

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

Step 2.1.3

Differentiate.

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Step 2.1.3.1

By the Sum Rule, the derivative of $\frac{x}{2} - 5$ with respect to $x$ is $\frac{d}{d x} [\frac{x}{2}] + \frac{d}{d x} [- 5]$ .

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

Step 2.1.3.2

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

$f' (x) = x (4 {(\frac{x}{2} - 5)}^{3} (\frac{1}{2} \cdot \frac{d}{d x} (x) + \frac{d}{d x} (- 5))) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.3

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

$f' (x) = x (4 {(\frac{x}{2} - 5)}^{3} (\frac{1}{2} \cdot 1 + \frac{d}{d x} (- 5))) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.4

Multiply $\frac{1}{2}$ by $1$ .

$f' (x) = x (4 {(\frac{x}{2} - 5)}^{3} (\frac{1}{2} + \frac{d}{d x} (- 5))) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.5

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

$f' (x) = x (4 {(\frac{x}{2} - 5)}^{3} (\frac{1}{2} + 0)) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.6

Simplify terms.

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Step 2.1.3.6.1

Add $\frac{1}{2}$ and $0$ .

$f' (x) = x (4 {(\frac{x}{2} - 5)}^{3} (\frac{1}{2})) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.6.2

Combine $\frac{1}{2}$ and $4$ .

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

Step 2.1.3.6.3

Combine $\frac{4}{2}$ and ${(\frac{x}{2} - 5)}^{3}$ .

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

Step 2.1.3.6.4

Cancel the common factor of $4$ and $2$ .

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Step 2.1.3.6.4.1

Factor $2$ out of $4 {(\frac{x}{2} - 5)}^{3}$ .

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

Step 2.1.3.6.4.2

Cancel the common factors.

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Step 2.1.3.6.4.2.1

Factor $2$ out of $2$ .

$f' (x) = x (\frac{2 (2 {(\frac{x}{2} - 5)}^{3})}{2 (1)}) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.6.4.2.2

Cancel the common factor.

$f' (x) = x (\frac{2 (2 {(\frac{x}{2} - 5)}^{3})}{2 \cdot 1}) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.6.4.2.3

Rewrite the expression.

$f' (x) = x (\frac{2 {(\frac{x}{2} - 5)}^{3}}{1}) + {(\frac{x}{2} - 5)}^{4} \frac{d}{d x} (x)$

Step 2.1.3.6.4.2.4

Divide $2 {(\frac{x}{2} - 5)}^{3}$ by $1$ .

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

Step 2.1.3.7

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

$f' (x) = x (2 {(\frac{x}{2} - 5)}^{3}) + {(\frac{x}{2} - 5)}^{4} \cdot 1$

Step 2.1.3.8

Multiply ${(\frac{x}{2} - 5)}^{4}$ by $1$ .

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

Step 2.1.4

Simplify.

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Step 2.1.4.1

Factor ${(\frac{x}{2} - 5)}^{3}$ out of $x (2 {(\frac{x}{2} - 5)}^{3}) + {(\frac{x}{2} - 5)}^{4}$ .

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Step 2.1.4.1.1

Reorder $x$ and $2$ .

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

Step 2.1.4.1.2

Factor ${(\frac{x}{2} - 5)}^{3}$ out of $2 \cdot x {(\frac{x}{2} - 5)}^{3}$ .

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

Step 2.1.4.1.3

Factor ${(\frac{x}{2} - 5)}^{3}$ out of ${(\frac{x}{2} - 5)}^{4}$ .

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

Step 2.1.4.1.4

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

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

Step 2.1.4.2

Combine terms.

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Step 2.1.4.2.1

To write $2 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.1.4.2.2

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

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

Step 2.1.4.2.3

Combine the numerators over the common denominator.

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

Step 2.1.4.2.4

Multiply $2$ by $2$ .

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

Step 2.1.4.2.5

Add $4 x$ and $x$ .

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

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = {(\frac{x}{2} - 5)}^{3}$ and $g (x) = \frac{5 x}{2} - 5$ .

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

Step 2.2.2

Differentiate.

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Step 2.2.2.1

By the Sum Rule, the derivative of $\frac{5 x}{2} - 5$ with respect to $x$ is $\frac{d}{d x} [\frac{5 x}{2}] + \frac{d}{d x} [- 5]$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

$f'' (x) = {(\frac{x}{2} - 5)}^{3} (\frac{5}{2} \cdot 1 + \frac{d}{d x} (- 5)) + (\frac{5 x}{2} - 5) \frac{d}{d x} ({(\frac{x}{2} - 5)}^{3})$

Step 2.2.2.4

Multiply $\frac{5}{2}$ by $1$ .

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

Step 2.2.2.5

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

$f'' (x) = {(\frac{x}{2} - 5)}^{3} (\frac{5}{2} + 0) + (\frac{5 x}{2} - 5) \frac{d}{d x} ({(\frac{x}{2} - 5)}^{3})$

Step 2.2.2.6

Combine fractions.

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Step 2.2.2.6.1

Add $\frac{5}{2}$ and $0$ .

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

Step 2.2.2.6.2

Combine ${(\frac{x}{2} - 5)}^{3}$ and $\frac{5}{2}$ .

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

Step 2.2.2.6.3

Move $5$ to the left of ${(\frac{x}{2} - 5)}^{3}$ .

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

Step 2.2.3

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 $g (x) = \frac{x}{2} - 5$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u$ as $\frac{x}{2} - 5$ .

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

Step 2.2.3.2

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

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

Step 2.2.3.3

Replace all occurrences of $u$ with $\frac{x}{2} - 5$ .

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

Step 2.2.4

Differentiate.

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Step 2.2.4.1

Move $3$ to the left of $\frac{5 x}{2} - 5$ .

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

Step 2.2.4.2

By the Sum Rule, the derivative of $\frac{x}{2} - 5$ with respect to $x$ is $\frac{d}{d x} [\frac{x}{2}] + \frac{d}{d x} [- 5]$ .

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

Step 2.2.4.3

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3}}{2} + 3 (\frac{5 x}{2} - 5) {(\frac{x}{2} - 5)}^{2} (\frac{1}{2} \cdot \frac{d}{d x} (x) + \frac{d}{d x} (- 5))$

Step 2.2.4.4

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

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3}}{2} + 3 (\frac{5 x}{2} - 5) {(\frac{x}{2} - 5)}^{2} (\frac{1}{2} \cdot 1 + \frac{d}{d x} (- 5))$

Step 2.2.4.5

Multiply $\frac{1}{2}$ by $1$ .

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3}}{2} + 3 (\frac{5 x}{2} - 5) {(\frac{x}{2} - 5)}^{2} (\frac{1}{2} + \frac{d}{d x} (- 5))$

Step 2.2.4.6

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

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3}}{2} + 3 (\frac{5 x}{2} - 5) {(\frac{x}{2} - 5)}^{2} (\frac{1}{2} + 0)$

Step 2.2.4.7

Combine fractions.

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Step 2.2.4.7.1

Add $\frac{1}{2}$ and $0$ .

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3}}{2} + 3 (\frac{5 x}{2} - 5) {(\frac{x}{2} - 5)}^{2} (\frac{1}{2})$

Step 2.2.4.7.2

Combine $\frac{1}{2}$ and $3$ .

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

Step 2.2.4.7.3

Combine ${(\frac{x}{2} - 5)}^{2}$ and $\frac{3}{2}$ .

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

Step 2.2.4.7.4

Move $3$ to the left of ${(\frac{x}{2} - 5)}^{2}$ .

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

Step 2.2.5

To write $\frac{3 {(\frac{x}{2} - 5)}^{2}}{2} (\frac{5 x}{2} - 5)$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 2.2.6

Combine $\frac{3 {(\frac{x}{2} - 5)}^{2}}{2} (\frac{5 x}{2} - 5)$ and $\frac{2}{2}$ .

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

Step 2.2.7

Combine the numerators over the common denominator.

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

Step 2.2.8

Combine $2$ and $\frac{3 {(\frac{x}{2} - 5)}^{2}}{2}$ .

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

Step 2.2.9

Multiply $3$ by $2$ .

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

Step 2.2.10

Cancel the common factor of $6$ and $2$ .

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Step 2.2.10.1

Factor $2$ out of $6 {(\frac{x}{2} - 5)}^{2}$ .

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

Step 2.2.10.2

Cancel the common factors.

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Step 2.2.10.2.1

Factor $2$ out of $2$ .

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3} + \frac{2 (3 {(\frac{x}{2} - 5)}^{2})}{2 (1)} \cdot (\frac{5 x}{2} - 5)}{2}$

Step 2.2.10.2.2

Cancel the common factor.

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3} + \frac{2 (3 {(\frac{x}{2} - 5)}^{2})}{2 \cdot 1} \cdot (\frac{5 x}{2} - 5)}{2}$

Step 2.2.10.2.3

Rewrite the expression.

$f'' (x) = \frac{5 {(\frac{x}{2} - 5)}^{3} + \frac{3 {(\frac{x}{2} - 5)}^{2}}{1} \cdot (\frac{5 x}{2} - 5)}{2}$

Step 2.2.10.2.4

Divide $3 {(\frac{x}{2} - 5)}^{2}$ by $1$ .

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

Step 2.2.11

Simplify.

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Step 2.2.11.1

Simplify the numerator.

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Step 2.2.11.1.1

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

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Step 2.2.11.1.1.1

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

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

Step 2.2.11.1.1.2

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

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

Step 2.2.11.1.1.3

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

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

Step 2.2.11.1.2

Apply the distributive property.

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

Step 2.2.11.1.3

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

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

Step 2.2.11.1.4

Multiply $5$ by $- 5$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x}{2} - 25 + 3 (\frac{5 x}{2} - 5))}{2}$

Step 2.2.11.1.5

Apply the distributive property.

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x}{2} - 25 + 3 (\frac{5 x}{2}) + 3 \cdot - 5)}{2}$

Step 2.2.11.1.6

Multiply $3 \frac{5 x}{2}$ .

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Step 2.2.11.1.6.1

Combine $3$ and $\frac{5 x}{2}$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x}{2} - 25 + \frac{3 (5 x)}{2} + 3 \cdot - 5)}{2}$

Step 2.2.11.1.6.2

Multiply $5$ by $3$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x}{2} - 25 + \frac{15 x}{2} + 3 \cdot - 5)}{2}$

Step 2.2.11.1.7

Multiply $3$ by $- 5$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x}{2} - 25 + \frac{15 x}{2} - 15)}{2}$

Step 2.2.11.1.8

Combine the numerators over the common denominator.

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x + 15 x}{2} - 25 - 15)}{2}$

Step 2.2.11.1.9

Subtract $15$ from $- 25$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5)}^{2} (\frac{5 x + 15 x}{2} - 40)}{2}$

Step 2.2.11.1.10

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f'' (x) = \frac{{(\frac{x}{2} - 5 \cdot \frac{2}{2})}^{2} (\frac{5 x + 15 x}{2} - 40)}{2}$

Step 2.2.11.1.11

Combine $- 5$ and $\frac{2}{2}$ .

$f'' (x) = \frac{{(\frac{x}{2} + \frac{- 5 \cdot 2}{2})}^{2} (\frac{5 x + 15 x}{2} - 40)}{2}$

Step 2.2.11.1.12

Combine the numerators over the common denominator.

$f'' (x) = \frac{{(\frac{x - 5 \cdot 2}{2})}^{2} (\frac{5 x + 15 x}{2} - 40)}{2}$

Step 2.2.11.1.13

Multiply $- 5$ by $2$ .

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{5 x + 15 x}{2} - 40)}{2}$

Step 2.2.11.1.14

Add $5 x$ and $15 x$ .

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{20 x}{2} - 40)}{2}$

Step 2.2.11.1.15

Cancel the common factor of $20$ and $2$ .

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Step 2.2.11.1.15.1

Factor $2$ out of $20 x$ .

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{2 (10 x)}{2} - 40)}{2}$

Step 2.2.11.1.15.2

Cancel the common factors.

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Step 2.2.11.1.15.2.1

Factor $2$ out of $2$ .

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{2 (10 x)}{2 (1)} - 40)}{2}$

Step 2.2.11.1.15.2.2

Cancel the common factor.

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{2 (10 x)}{2 \cdot 1} - 40)}{2}$

Step 2.2.11.1.15.2.3

Rewrite the expression.

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (\frac{10 x}{1} - 40)}{2}$

Step 2.2.11.1.15.2.4

Divide $10 x$ by $1$ .

$f'' (x) = \frac{{(\frac{x - 10}{2})}^{2} (10 x - 40)}{2}$

Step 2.2.11.1.16

Apply the product rule to $\frac{x - 10}{2}$ .

$f'' (x) = \frac{\frac{{(x - 10)}^{2}}{2^{2}} \cdot (10 x - 40)}{2}$

Step 2.2.11.1.17

Factor $10$ out of $10 x - 40$ .

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Step 2.2.11.1.17.1

Factor $10$ out of $10 x$ .

$f'' (x) = \frac{\frac{{(x - 10)}^{2}}{2^{2}} \cdot (10 (x) - 40)}{2}$

Step 2.2.11.1.17.2

Factor $10$ out of $- 40$ .

$f'' (x) = \frac{\frac{{(x - 10)}^{2}}{2^{2}} \cdot (10 x + 10 \cdot - 4)}{2}$

Step 2.2.11.1.17.3

Factor $10$ out of $10 x + 10 \cdot - 4$ .

$f'' (x) = \frac{\frac{{(x - 10)}^{2}}{2^{2}} \cdot (10 (x - 4))}{2}$

Step 2.2.11.1.18

Combine $\frac{{(x - 10)}^{2}}{2^{2}}$ and $10$ .

$f'' (x) = \frac{\frac{{(x - 10)}^{2} \cdot 10}{2^{2}} \cdot (x - 4)}{2}$

Step 2.2.11.1.19

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

$f'' (x) = \frac{\frac{{(x - 10)}^{2} \cdot 10}{4} \cdot (x - 4)}{2}$

Step 2.2.11.1.20

Cancel the common factor of $10$ and $4$ .

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Step 2.2.11.1.20.1

Factor $2$ out of ${(x - 10)}^{2} \cdot 10$ .

$f'' (x) = \frac{\frac{2 ({(x - 10)}^{2} \cdot 5)}{4} \cdot (x - 4)}{2}$

Step 2.2.11.1.20.2

Cancel the common factors.

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Step 2.2.11.1.20.2.1

Factor $2$ out of $4$ .

$f'' (x) = \frac{\frac{2 ({(x - 10)}^{2} \cdot 5)}{2 (2)} \cdot (x - 4)}{2}$

Step 2.2.11.1.20.2.2

Cancel the common factor.

$f'' (x) = \frac{\frac{2 ({(x - 10)}^{2} \cdot 5)}{2 \cdot 2} \cdot (x - 4)}{2}$

Step 2.2.11.1.20.2.3

Rewrite the expression.

$f'' (x) = \frac{\frac{{(x - 10)}^{2} \cdot 5}{2} \cdot (x - 4)}{2}$

Step 2.2.11.1.21

Move $5$ to the left of ${(x - 10)}^{2}$ .

$f'' (x) = \frac{\frac{5 {(x - 10)}^{2}}{2} \cdot (x - 4)}{2}$

Step 2.2.11.2

Reorder terms.

$f'' (x) = \frac{(x - 4) (\frac{5 {(x - 10)}^{2}}{2})}{2}$

Step 2.2.11.3

Factor $\frac{5 {(x - 10)}^{2}}{2}$ out of $\frac{(x - 4) \frac{5 {(x - 10)}^{2}}{2}}{2}$ .

$f'' (x) = \frac{5 {(x - 10)}^{2}}{2} \cdot \frac{x - 4}{2}$

Step 2.2.11.4

Multiply $\frac{5 {(x - 10)}^{2}}{2} \cdot \frac{x - 4}{2}$ .

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Step 2.2.11.4.1

Multiply $\frac{5 {(x - 10)}^{2}}{2}$ by $\frac{x - 4}{2}$ .

$f'' (x) = \frac{5 {(x - 10)}^{2} (x - 4)}{2 \cdot 2}$

Step 2.2.11.4.2

Multiply $2$ by $2$ .

$f'' (x) = \frac{5 {(x - 10)}^{2} (x - 4)}{4}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{5 {(x - 10)}^{2} (x - 4)}{4}$ .

$\frac{5 {(x - 10)}^{2} (x - 4)}{4}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{5 {(x - 10)}^{2} (x - 4)}{4} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$\frac{5 {(x - 10)}^{2} (x - 4)}{4} = 0$

Step 3.2

Set the numerator equal to zero.

$5 {(x - 10)}^{2} (x - 4) = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x - 10)}^{2} = 0$

$x - 4 = 0$

Step 3.3.2

Set ${(x - 10)}^{2}$ equal to $0$ and solve for $x$ .

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Step 3.3.2.1

Set ${(x - 10)}^{2}$ equal to $0$ .

${(x - 10)}^{2} = 0$

Step 3.3.2.2

Solve ${(x - 10)}^{2} = 0$ for $x$ .

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Step 3.3.2.2.1

Set the $x - 10$ equal to $0$ .

$x - 10 = 0$

Step 3.3.2.2.2

Add $10$ to both sides of the equation.

$x = 10$

Step 3.3.3

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 3.3.3.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 3.3.3.2

Add $4$ to both sides of the equation.

$x = 4$

Step 3.3.4

The final solution is all the values that make $5 {(x - 10)}^{2} (x - 4) = 0$ true.

$x = 10, 4$

Step 4

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

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Step 4.1

Substitute $10$ in $f (x) = x {(\frac{x}{2} - 5)}^{4}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $10$ in the expression.

$f (10) = (10) {(\frac{10}{2} - 5)}^{4}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Divide $10$ by $2$ .

$f (10) = 10 {(5 - 5)}^{4}$

Step 4.1.2.2

Subtract $5$ from $5$ .

$f (10) = 10 \cdot 0^{4}$

Step 4.1.2.3

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

$f (10) = 10 \cdot 0$

Step 4.1.2.4

Multiply $10$ by $0$ .

$f (10) = 0$

Step 4.1.2.5

The final answer is $0$ .

$0$

Step 4.2

The point found by substituting $10$ in $f (x) = x {(\frac{x}{2} - 5)}^{4}$ is $(10, 0)$ . This point can be an inflection point.

$(10, 0)$

Step 4.3

Substitute $4$ in $f (x) = x {(\frac{x}{2} - 5)}^{4}$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $4$ in the expression.

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

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Divide $4$ by $2$ .

$f (4) = 4 {(2 - 5)}^{4}$

Step 4.3.2.2

Subtract $5$ from $2$ .

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

Step 4.3.2.3

Raise $- 3$ to the power of $4$ .

$f (4) = 4 \cdot 81$

Step 4.3.2.4

Multiply $4$ by $81$ .

$f (4) = 324$

Step 4.3.2.5

The final answer is $324$ .

$324$

Step 4.4

The point found by substituting $4$ in $f (x) = x {(\frac{x}{2} - 5)}^{4}$ is $(4, 324)$ . This point can be an inflection point.

$(4, 324)$

Step 4.5

Determine the points that could be inflection points.

$(10, 0), (4, 324)$

Step 5

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

$(- \infty, 4) \cup (4, 10) \cup (10, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 4)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $3.9$ in the expression.

$f'' (3.9) = \frac{5 {((3.9) - 10)}^{2} ((3.9) - 4)}{4}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify the numerator.

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Step 6.2.1.1

Subtract $10$ from $3.9$ .

$f'' (3.9) = \frac{5 {(- 6.1)}^{2} (3.9 - 4)}{4}$

Step 6.2.1.2

Subtract $4$ from $3.9$ .

$f'' (3.9) = \frac{5 {(- 6.1)}^{2} \cdot - 0.1}{4}$

Step 6.2.1.3

Combine exponents.

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Step 6.2.1.3.1

Multiply $- 0.1$ by $5$ .

$f'' (3.9) = \frac{- 0.5 {(- 6.1)}^{2}}{4}$

Step 6.2.1.3.2

Multiply $- 0.5$ by ${(- 6.1)}^{2}$ .

$f'' (3.9) = \frac{- 18.605}{4}$

Step 6.2.2

Divide $- 18.605$ by $4$ .

$f'' (3.9) = - 4.65125$

Step 6.2.3

The final answer is $- 4.65125$ .

$- 4.65125$

Step 6.3

At $3.9$ , the second derivative is $- 4.65125$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, 4)$

Decreasing on $(- \infty, 4)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(4, 10)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $7$ in the expression.

$f'' (7) = \frac{5 {((7) - 10)}^{2} ((7) - 4)}{4}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the numerator.

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Step 7.2.1.1

Subtract $10$ from $7$ .

$f'' (7) = \frac{5 {(- 3)}^{2} (7 - 4)}{4}$

Step 7.2.1.2

Subtract $4$ from $7$ .

$f'' (7) = \frac{5 {(- 3)}^{2} \cdot 3}{4}$

Step 7.2.1.3

Multiply $3$ by $5$ .

$f'' (7) = \frac{15 {(- 3)}^{2}}{4}$

Step 7.2.1.4

Raise $- 3$ to the power of $2$ .

$f'' (7) = \frac{15 \cdot 9}{4}$

Step 7.2.2

Multiply $15$ by $9$ .

$f'' (7) = \frac{135}{4}$

Step 7.2.3

The final answer is $\frac{135}{4}$ .

$\frac{135}{4}$

Step 7.3

At $7$ , the second derivative is $33.75$ . Since this is positive, the second derivative is increasing on the interval $(4, 10)$ .

Increasing on $(4, 10)$ since $f'' (x) > 0$

Step 8

Substitute a value from the interval $(10, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $10.1$ in the expression.

$f'' (10.1) = \frac{5 {((10.1) - 10)}^{2} ((10.1) - 4)}{4}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the numerator.

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Step 8.2.1.1

Subtract $10$ from $10.1$ .

$f'' (10.1) = \frac{5 \cdot ({0.1}^{2} (10.1 - 4))}{4}$

Step 8.2.1.2

Subtract $4$ from $10.1$ .

$f'' (10.1) = \frac{5 \cdot {0.1}^{2} \cdot 6.1}{4}$

Step 8.2.1.3

Combine exponents.

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Step 8.2.1.3.1

Multiply $6.1$ by $5$ .

$f'' (10.1) = \frac{30.5 \cdot {0.1}^{2}}{4}$

Step 8.2.1.3.2

Multiply $30.5$ by ${0.1}^{2}$ .

$f'' (10.1) = \frac{0.305}{4}$

Step 8.2.2

Divide $0.305$ by $4$ .

$f'' (10.1) = 0.07625$

Step 8.2.3

The final answer is $0.07625$ .

$0.07625$

Step 8.3

At $10.1$ , the second derivative is $0.07625$ . Since this is positive, the second derivative is increasing on the interval $(10, \infty)$ .

Increasing on $(10, \infty)$ since $f'' (x) > 0$

Step 9

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 $(4, 324)$ .

$(4, 324)$

Step 10