Calculus Examples

$f (x) = 100 - \frac{225}{x + 5} - x$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $100 - \frac{225}{x + 5} - x$ with respect to $x$ is $\frac{d}{d x} [100] + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [100] + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$

Step 1.1.2

Since $100$ is constant with respect to $x$ , the derivative of $100$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$

Step 1.2

Evaluate $\frac{d}{d x} [- \frac{225}{x + 5}]$ .

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Step 1.2.1

Since $- 225$ is constant with respect to $x$ , the derivative of $- \frac{225}{x + 5}$ with respect to $x$ is $- 225 \frac{d}{d x} [\frac{1}{x + 5}]$ .

$0 - 225 \frac{d}{d x} [\frac{1}{x + 5}] + \frac{d}{d x} [- x]$

Step 1.2.2

Rewrite $\frac{1}{x + 5}$ as ${(x + 5)}^{- 1}$ .

$0 - 225 \frac{d}{d x} [{(x + 5)}^{- 1}] + \frac{d}{d x} [- x]$

Step 1.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x + 5$ .

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Step 1.2.3.1

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

$0 - 225 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 1.2.3.2

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

$0 - 225 (- u^{- 2} \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 1.2.3.3

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

$0 - 225 (- {(x + 5)}^{- 2} \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 1.2.4

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

$0 - 225 (- {(x + 5)}^{- 2} (\frac{d}{d x} [x] + \frac{d}{d x} [5])) + \frac{d}{d x} [- x]$

Step 1.2.5

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

$0 - 225 (- {(x + 5)}^{- 2} (1 + \frac{d}{d x} [5])) + \frac{d}{d x} [- x]$

Step 1.2.6

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

$0 - 225 (- {(x + 5)}^{- 2} (1 + 0)) + \frac{d}{d x} [- x]$

Step 1.2.7

Add $1$ and $0$ .

$0 - 225 (- {(x + 5)}^{- 2} \cdot 1) + \frac{d}{d x} [- x]$

Step 1.2.8

Multiply $- 1$ by $1$ .

$0 - 225 (- {(x + 5)}^{- 2}) + \frac{d}{d x} [- x]$

Step 1.2.9

Multiply $- 1$ by $- 225$ .

$0 + 225 {(x + 5)}^{- 2} + \frac{d}{d x} [- x]$

Step 1.3

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

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Step 1.3.1

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

$0 + 225 {(x + 5)}^{- 2} - \frac{d}{d x} [x]$

Step 1.3.2

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

$0 + 225 {(x + 5)}^{- 2} - 1 \cdot 1$

Step 1.3.3

Multiply $- 1$ by $1$ .

$0 + 225 {(x + 5)}^{- 2} - 1$

Step 1.4

Simplify.

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Step 1.4.1

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

$0 + 225 \frac{1}{{(x + 5)}^{2}} - 1$

Step 1.4.2

Combine terms.

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Step 1.4.2.1

Combine $225$ and $\frac{1}{{(x + 5)}^{2}}$ .

$0 + \frac{225}{{(x + 5)}^{2}} - 1$

Step 1.4.2.2

Add $0$ and $\frac{225}{{(x + 5)}^{2}}$ .

$\frac{225}{{(x + 5)}^{2}} - 1$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $\frac{225}{{(x + 5)}^{2}} - 1$ with respect to $x$ is $\frac{d}{d x} [\frac{225}{{(x + 5)}^{2}}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (\frac{225}{{(x + 5)}^{2}}) + \frac{d}{d x} (- 1)$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{225}{{(x + 5)}^{2}}]$ .

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Step 2.2.1

Since $225$ is constant with respect to $x$ , the derivative of $\frac{225}{{(x + 5)}^{2}}$ with respect to $x$ is $225 \frac{d}{d x} [\frac{1}{{(x + 5)}^{2}}]$ .

$f'' (x) = 225 \frac{d}{d x} (\frac{1}{{(x + 5)}^{2}}) + \frac{d}{d x} (- 1)$

Step 2.2.2

Rewrite $\frac{1}{{(x + 5)}^{2}}$ as ${({(x + 5)}^{2})}^{- 1}$ .

$f'' (x) = 225 \frac{d}{d x} ({({(x + 5)}^{2})}^{- 1}) + \frac{d}{d x} (- 1)$

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 $f (x) = x^{- 1}$ and $g (x) = {(x + 5)}^{2}$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as ${(x + 5)}^{2}$ .

$f'' (x) = 225 (\frac{d}{d u (1)} ({u_{1}}^{- 1}) \frac{d}{d x} ({(x + 5)}^{2})) + \frac{d}{d x} (- 1)$

Step 2.2.3.2

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

$f'' (x) = 225 (- {u_{1}}^{- 2} \frac{d}{d x} {(x + 5)}^{2}) + \frac{d}{d x} (- 1)$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with ${(x + 5)}^{2}$ .

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} \frac{d}{d x} {(x + 5)}^{2}) + \frac{d}{d x} (- 1)$

Step 2.2.4

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

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Step 2.2.4.1

To apply the Chain Rule, set $u_{2}$ as $x + 5$ .

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (\frac{d}{d u (2)} ({u_{2}}^{2}) \frac{d}{d x} (x + 5))) + \frac{d}{d x} (- 1)$

Step 2.2.4.2

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

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (2 u_{2} \frac{d}{d x} (x + 5))) + \frac{d}{d x} (- 1)$

Step 2.2.4.3

Replace all occurrences of $u_{2}$ with $x + 5$ .

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (2 (x + 5) \frac{d}{d x} (x + 5))) + \frac{d}{d x} (- 1)$

Step 2.2.5

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

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (2 (x + 5) (\frac{d}{d x} (x) + \frac{d}{d x} (5)))) + \frac{d}{d x} (- 1)$

Step 2.2.6

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

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (2 (x + 5) (1 + \frac{d}{d x} (5)))) + \frac{d}{d x} (- 1)$

Step 2.2.7

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

$f'' (x) = 225 (- {({(x + 5)}^{2})}^{- 2} (2 (x + 5) (1 + 0))) + \frac{d}{d x} (- 1)$

Step 2.2.8

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

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Step 2.2.8.1

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

$f'' (x) = 225 (- {(x + 5)}^{2 \cdot - 2} (2 (x + 5) (1 + 0))) + \frac{d}{d x} (- 1)$

Step 2.2.8.2

Multiply $2$ by $- 2$ .

$f'' (x) = 225 (- {(x + 5)}^{- 4} (2 (x + 5) (1 + 0))) + \frac{d}{d x} (- 1)$

Step 2.2.9

Add $1$ and $0$ .

$f'' (x) = 225 (- {(x + 5)}^{- 4} (2 (x + 5) \cdot 1)) + \frac{d}{d x} (- 1)$

Step 2.2.10

Multiply $2$ by $1$ .

$f'' (x) = 225 (- {(x + 5)}^{- 4} (2 (x + 5))) + \frac{d}{d x} (- 1)$

Step 2.2.11

Multiply $2$ by $- 1$ .

$f'' (x) = 225 (- 2 {(x + 5)}^{- 4} (x + 5)) + \frac{d}{d x} (- 1)$

Step 2.2.12

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

$f'' (x) = 225 (- 2 ((x + 5) {(x + 5)}^{- 4})) + \frac{d}{d x} (- 1)$

Step 2.2.13

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

$f'' (x) = 225 (- 2 {(x + 5)}^{1 - 4}) + \frac{d}{d x} (- 1)$

Step 2.2.14

Subtract $4$ from $1$ .

$f'' (x) = 225 (- 2 {(x + 5)}^{- 3}) + \frac{d}{d x} (- 1)$

Step 2.2.15

Multiply $- 2$ by $225$ .

$f'' (x) = - 450 {(x + 5)}^{- 3} + \frac{d}{d x} (- 1)$

Step 2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = - 450 {(x + 5)}^{- 3} + 0$

Step 2.4

Simplify.

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Step 2.4.1

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

$f'' (x) = - 450 \frac{1}{{(x + 5)}^{3}} + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Combine $- 450$ and $\frac{1}{{(x + 5)}^{3}}$ .

$f'' (x) = \frac{- 450}{{(x + 5)}^{3}} + 0$

Step 2.4.2.2

Move the negative in front of the fraction.

$f'' (x) = - \frac{450}{{(x + 5)}^{3}} + 0$

Step 2.4.2.3

Add $- \frac{450}{{(x + 5)}^{3}}$ and $0$ .

$f'' (x) = - \frac{450}{{(x + 5)}^{3}}$

Step 3

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

$\frac{225}{{(x + 5)}^{2}} - 1 = 0$

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

Differentiate.

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Step 4.1.1.1

By the Sum Rule, the derivative of $100 - \frac{225}{x + 5} - x$ with respect to $x$ is $\frac{d}{d x} [100] + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [100] + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$

Step 4.1.1.2

Since $100$ is constant with respect to $x$ , the derivative of $100$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [- \frac{225}{x + 5}] + \frac{d}{d x} [- x]$

Step 4.1.2

Evaluate $\frac{d}{d x} [- \frac{225}{x + 5}]$ .

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Step 4.1.2.1

Since $- 225$ is constant with respect to $x$ , the derivative of $- \frac{225}{x + 5}$ with respect to $x$ is $- 225 \frac{d}{d x} [\frac{1}{x + 5}]$ .

$0 - 225 \frac{d}{d x} [\frac{1}{x + 5}] + \frac{d}{d x} [- x]$

Step 4.1.2.2

Rewrite $\frac{1}{x + 5}$ as ${(x + 5)}^{- 1}$ .

$0 - 225 \frac{d}{d x} [{(x + 5)}^{- 1}] + \frac{d}{d x} [- x]$

Step 4.1.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x + 5$ .

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Step 4.1.2.3.1

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

$0 - 225 (\frac{d}{d u} [u^{- 1}] \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 4.1.2.3.2

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

$0 - 225 (- u^{- 2} \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 4.1.2.3.3

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

$0 - 225 (- {(x + 5)}^{- 2} \frac{d}{d x} [x + 5]) + \frac{d}{d x} [- x]$

Step 4.1.2.4

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

$0 - 225 (- {(x + 5)}^{- 2} (\frac{d}{d x} [x] + \frac{d}{d x} [5])) + \frac{d}{d x} [- x]$

Step 4.1.2.5

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

$0 - 225 (- {(x + 5)}^{- 2} (1 + \frac{d}{d x} [5])) + \frac{d}{d x} [- x]$

Step 4.1.2.6

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

$0 - 225 (- {(x + 5)}^{- 2} (1 + 0)) + \frac{d}{d x} [- x]$

Step 4.1.2.7

Add $1$ and $0$ .

$0 - 225 (- {(x + 5)}^{- 2} \cdot 1) + \frac{d}{d x} [- x]$

Step 4.1.2.8

Multiply $- 1$ by $1$ .

$0 - 225 (- {(x + 5)}^{- 2}) + \frac{d}{d x} [- x]$

Step 4.1.2.9

Multiply $- 1$ by $- 225$ .

$0 + 225 {(x + 5)}^{- 2} + \frac{d}{d x} [- x]$

Step 4.1.3

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

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Step 4.1.3.1

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

$0 + 225 {(x + 5)}^{- 2} - \frac{d}{d x} [x]$

Step 4.1.3.2

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

$0 + 225 {(x + 5)}^{- 2} - 1 \cdot 1$

Step 4.1.3.3

Multiply $- 1$ by $1$ .

$0 + 225 {(x + 5)}^{- 2} - 1$

Step 4.1.4

Simplify.

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Step 4.1.4.1

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

$0 + 225 \frac{1}{{(x + 5)}^{2}} - 1$

Step 4.1.4.2

Combine terms.

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Step 4.1.4.2.1

Combine $225$ and $\frac{1}{{(x + 5)}^{2}}$ .

$0 + \frac{225}{{(x + 5)}^{2}} - 1$

Step 4.1.4.2.2

Add $0$ and $\frac{225}{{(x + 5)}^{2}}$ .

$f' (x) = \frac{225}{{(x + 5)}^{2}} - 1$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{225}{{(x + 5)}^{2}} - 1$ .

$\frac{225}{{(x + 5)}^{2}} - 1$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{225}{{(x + 5)}^{2}} - 1 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{225}{{(x + 5)}^{2}} - 1 = 0$

Step 5.2

Add $1$ to both sides of the equation.

$\frac{225}{{(x + 5)}^{2}} = 1$

Step 5.3

Find the LCD of the terms in the equation.

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Step 5.3.1

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

${(x + 5)}^{2}, 1$

Step 5.3.2

The LCM of one and any expression is the expression.

${(x + 5)}^{2}$

Step 5.4

Multiply each term in $\frac{225}{{(x + 5)}^{2}} = 1$ by ${(x + 5)}^{2}$ to eliminate the fractions.

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Step 5.4.1

Multiply each term in $\frac{225}{{(x + 5)}^{2}} = 1$ by ${(x + 5)}^{2}$ .

$\frac{225}{{(x + 5)}^{2}} {(x + 5)}^{2} = 1 {(x + 5)}^{2}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of ${(x + 5)}^{2}$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{225}{{(x + 5)}^{2}} {(x + 5)}^{2} = 1 {(x + 5)}^{2}$

Step 5.4.2.1.2

Rewrite the expression.

$225 = 1 {(x + 5)}^{2}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Multiply ${(x + 5)}^{2}$ by $1$ .

$225 = {(x + 5)}^{2}$

Step 5.5

Solve the equation.

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Step 5.5.1

Rewrite the equation as ${(x + 5)}^{2} = 225$ .

${(x + 5)}^{2} = 225$

Step 5.5.2

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

$x + 5 = \pm \sqrt{225}$

Step 5.5.3

Simplify $\pm \sqrt{225}$ .

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Step 5.5.3.1

Rewrite $225$ as $15^{2}$ .

$x + 5 = \pm \sqrt{15^{2}}$

Step 5.5.3.2

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

$x + 5 = \pm 15$

Step 5.5.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.5.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x + 5 = 15$

Step 5.5.4.2

Move all terms not containing $x$ to the right side of the equation.

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Step 5.5.4.2.1

Subtract $5$ from both sides of the equation.

$x = 15 - 5$

Step 5.5.4.2.2

Subtract $5$ from $15$ .

$x = 10$

Step 5.5.4.3

Next, use the negative value of the $\pm$ to find the second solution.

$x + 5 = - 15$

Step 5.5.4.4

Move all terms not containing $x$ to the right side of the equation.

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Step 5.5.4.4.1

Subtract $5$ from both sides of the equation.

$x = - 15 - 5$

Step 5.5.4.4.2

Subtract $5$ from $- 15$ .

$x = - 20$

Step 5.5.4.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = 10, - 20$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the denominator in $\frac{225}{{(x + 5)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(x + 5)}^{2} = 0$

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.

$x = 10, - 20$

Step 8

Evaluate the second derivative at $x = 10$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{450}{{((10) + 5)}^{3}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify the denominator.

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Step 9.1.1

Add $10$ and $5$ .

$- \frac{450}{15^{3}}$

Step 9.1.2

Raise $15$ to the power of $3$ .

$- \frac{450}{3375}$

Step 9.2

Cancel the common factor of $450$ and $3375$ .

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Step 9.2.1

Factor $225$ out of $450$ .

$- \frac{225 (2)}{3375}$

Step 9.2.2

Cancel the common factors.

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Step 9.2.2.1

Factor $225$ out of $3375$ .

$- \frac{225 \cdot 2}{225 \cdot 15}$

Step 9.2.2.2

Cancel the common factor.

$- \frac{225 \cdot 2}{225 \cdot 15}$

Step 9.2.2.3

Rewrite the expression.

$- \frac{2}{15}$

Step 10

$x = 10$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 10$ is a local maximum

Step 11

Find the y-value when $x = 10$ .

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Step 11.1

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

$f (10) = 100 - \frac{225}{(10) + 5} - (10)$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Cancel the common factor of $225$ and $(10) + 5$ .

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Step 11.2.1.1.1

Factor $5$ out of $225$ .

$f (10) = 100 - \frac{5 \cdot 45}{10 + 5} - (10)$

Step 11.2.1.1.2

Cancel the common factors.

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Step 11.2.1.1.2.1

Factor $5$ out of $10$ .

$f (10) = 100 - \frac{5 \cdot 45}{5 (2) + 5} - (10)$

Step 11.2.1.1.2.2

Factor $5$ out of $5$ .

$f (10) = 100 - \frac{5 \cdot 45}{5 \cdot 2 + 5 \cdot 1} - (10)$

Step 11.2.1.1.2.3

Factor $5$ out of $5 \cdot 2 + 5 \cdot 1$ .

$f (10) = 100 - \frac{5 \cdot 45}{5 \cdot (2 + 1)} - (10)$

Step 11.2.1.1.2.4

Cancel the common factor.

$f (10) = 100 - \frac{5 \cdot 45}{5 \cdot (2 + 1)} - (10)$

Step 11.2.1.1.2.5

Rewrite the expression.

$f (10) = 100 - \frac{45}{2 + 1} - (10)$

Step 11.2.1.2

Add $2$ and $1$ .

$f (10) = 100 - \frac{45}{3} - (10)$

Step 11.2.1.3

Divide $45$ by $3$ .

$f (10) = 100 - 1 \cdot 15 - (10)$

Step 11.2.1.4

Multiply $- 1$ by $15$ .

$f (10) = 100 - 15 - (10)$

Step 11.2.1.5

Multiply $- 1$ by $10$ .

$f (10) = 100 - 15 - 10$

Step 11.2.2

Simplify by subtracting numbers.

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Step 11.2.2.1

Subtract $15$ from $100$ .

$f (10) = 85 - 10$

Step 11.2.2.2

Subtract $10$ from $85$ .

$f (10) = 75$

Step 11.2.3

The final answer is $75$ .

$y = 75$

Step 12

Evaluate the second derivative at $x = - 20$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{450}{{((- 20) + 5)}^{3}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify the denominator.

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Step 13.1.1

Add $- 20$ and $5$ .

$- \frac{450}{{(- 15)}^{3}}$

Step 13.1.2

Raise $- 15$ to the power of $3$ .

$- \frac{450}{- 3375}$

Step 13.2

Reduce the expression by cancelling the common factors.

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Step 13.2.1

Cancel the common factor of $450$ and $- 3375$ .

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Step 13.2.1.1

Factor $225$ out of $450$ .

$- \frac{225 (2)}{- 3375}$

Step 13.2.1.2

Cancel the common factors.

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Step 13.2.1.2.1

Factor $225$ out of $- 3375$ .

$- \frac{225 \cdot 2}{225 \cdot - 15}$

Step 13.2.1.2.2

Cancel the common factor.

$- \frac{225 \cdot 2}{225 \cdot - 15}$

Step 13.2.1.2.3

Rewrite the expression.

$- \frac{2}{- 15}$

Step 13.2.2

Move the negative in front of the fraction.

$- - \frac{2}{15}$

Step 13.3

Multiply $- - \frac{2}{15}$ .

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Step 13.3.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{2}{15})$

Step 13.3.2

Multiply $\frac{2}{15}$ by $1$ .

$\frac{2}{15}$

Step 14

$x = - 20$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 20$ is a local minimum

Step 15

Find the y-value when $x = - 20$ .

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Step 15.1

Replace the variable $x$ with $- 20$ in the expression.

$f (- 20) = 100 - \frac{225}{(- 20) + 5} - (- 20)$

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

Cancel the common factor of $225$ and $(- 20) + 5$ .

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Step 15.2.1.1.1

Factor $5$ out of $225$ .

$f (- 20) = 100 - \frac{5 \cdot 45}{- 20 + 5} - (- 20)$

Step 15.2.1.1.2

Cancel the common factors.

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Step 15.2.1.1.2.1

Factor $5$ out of $- 20$ .

$f (- 20) = 100 - \frac{5 \cdot 45}{5 (- 4) + 5} - (- 20)$

Step 15.2.1.1.2.2

Factor $5$ out of $5$ .

$f (- 20) = 100 - \frac{5 \cdot 45}{5 \cdot - 4 + 5 \cdot 1} - (- 20)$

Step 15.2.1.1.2.3

Factor $5$ out of $5 \cdot - 4 + 5 \cdot 1$ .

$f (- 20) = 100 - \frac{5 \cdot 45}{5 \cdot (- 4 + 1)} - (- 20)$

Step 15.2.1.1.2.4

Cancel the common factor.

$f (- 20) = 100 - \frac{5 \cdot 45}{5 \cdot (- 4 + 1)} - (- 20)$

Step 15.2.1.1.2.5

Rewrite the expression.

$f (- 20) = 100 - \frac{45}{- 4 + 1} - (- 20)$

Step 15.2.1.2

Add $- 4$ and $1$ .

$f (- 20) = 100 - \frac{45}{- 3} - (- 20)$

Step 15.2.1.3

Divide $45$ by $- 3$ .

$f (- 20) = 100 + 15 - (- 20)$

Step 15.2.1.4

Multiply $- 1$ by $- 15$ .

$f (- 20) = 100 + 15 - (- 20)$

Step 15.2.1.5

Multiply $- 1$ by $- 20$ .

$f (- 20) = 100 + 15 + 20$

Step 15.2.2

Simplify by adding numbers.

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Step 15.2.2.1

Add $100$ and $15$ .

$f (- 20) = 115 + 20$

Step 15.2.2.2

Add $115$ and $20$ .

$f (- 20) = 135$

Step 15.2.3

The final answer is $135$ .

$y = 135$

Step 16

These are the local extrema for $f (x) = 100 - \frac{225}{x + 5} - x$ .

$(10, 75)$ is a local maxima

$(- 20, 135)$ is a local minima

Step 17