Calculus Examples

$f (x) = \ln (x^{2} + 5 x + 9)$ , $[- 3, 1]$

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 using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x^{2} + 5 x + 9$ .

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Step 1.1.1.1.1

To apply the Chain Rule, set $u$ as $x^{2} + 5 x + 9$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} + 5 x + 9]$

Step 1.1.1.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [x^{2} + 5 x + 9]$

Step 1.1.1.1.3

Replace all occurrences of $u$ with $x^{2} + 5 x + 9$ .

$\frac{1}{x^{2} + 5 x + 9} \frac{d}{d x} [x^{2} + 5 x + 9]$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $x^{2} + 5 x + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [9]$ .

$\frac{1}{x^{2} + 5 x + 9} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x] + \frac{d}{d x} [9])$

Step 1.1.1.2.2

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

$\frac{1}{x^{2} + 5 x + 9} (2 x + \frac{d}{d x} [5 x] + \frac{d}{d x} [9])$

Step 1.1.1.2.3

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

$\frac{1}{x^{2} + 5 x + 9} (2 x + 5 \frac{d}{d x} [x] + \frac{d}{d x} [9])$

Step 1.1.1.2.4

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

$\frac{1}{x^{2} + 5 x + 9} (2 x + 5 \cdot 1 + \frac{d}{d x} [9])$

Step 1.1.1.2.5

Multiply $5$ by $1$ .

$\frac{1}{x^{2} + 5 x + 9} (2 x + 5 + \frac{d}{d x} [9])$

Step 1.1.1.2.6

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$\frac{1}{x^{2} + 5 x + 9} (2 x + 5 + 0)$

Step 1.1.1.2.7

Add $2 x + 5$ and $0$ .

$\frac{1}{x^{2} + 5 x + 9} (2 x + 5)$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Reorder the factors of $\frac{1}{x^{2} + 5 x + 9} (2 x + 5)$ .

$(2 x + 5) \frac{1}{x^{2} + 5 x + 9}$

Step 1.1.1.3.2

Multiply $2 x + 5$ by $\frac{1}{x^{2} + 5 x + 9}$ .

$f' (x) = \frac{2 x + 5}{x^{2} + 5 x + 9}$

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{2 x + 5}{x^{2} + 5 x + 9}$ .

$\frac{2 x + 5}{x^{2} + 5 x + 9}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\frac{2 x + 5}{x^{2} + 5 x + 9} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\frac{2 x + 5}{x^{2} + 5 x + 9} = 0$

Step 1.2.2

Set the numerator equal to zero.

$2 x + 5 = 0$

Step 1.2.3

Solve the equation for $x$ .

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Step 1.2.3.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 1.2.3.2

Divide each term in $2 x = - 5$ by $2$ and simplify.

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Step 1.2.3.2.1

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 1.2.3.2.2

Simplify the left side.

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Step 1.2.3.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.3.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Step 1.2.3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Step 1.2.3.2.3

Simplify the right side.

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Step 1.2.3.2.3.1

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 1.4

Evaluate $\ln (x^{2} + 5 x + 9)$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = - \frac{5}{2}$ .

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Step 1.4.1.1

Substitute $- \frac{5}{2}$ for $x$ .

$\ln ({(- \frac{5}{2})}^{2} + 5 (- \frac{5}{2}) + 9)$

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

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 1.4.1.2.1.1.1

Apply the product rule to $- \frac{5}{2}$ .

$\ln ({(- 1)}^{2} {(\frac{5}{2})}^{2} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.1.2

Apply the product rule to $\frac{5}{2}$ .

$\ln ({(- 1)}^{2} \frac{5^{2}}{2^{2}} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.2

Raise $- 1$ to the power of $2$ .

$\ln (1 \frac{5^{2}}{2^{2}} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.3

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

$\ln (\frac{5^{2}}{2^{2}} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.4

Raise $5$ to the power of $2$ .

$\ln (\frac{25}{2^{2}} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.5

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

$\ln (\frac{25}{4} + 5 (- \frac{5}{2}) + 9)$

Step 1.4.1.2.1.6

Multiply $5 (- \frac{5}{2})$ .

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Step 1.4.1.2.1.6.1

Multiply $- 1$ by $5$ .

$\ln (\frac{25}{4} - 5 (\frac{5}{2}) + 9)$

Step 1.4.1.2.1.6.2

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

$\ln (\frac{25}{4} + \frac{- 5 \cdot 5}{2} + 9)$

Step 1.4.1.2.1.6.3

Multiply $- 5$ by $5$ .

$\ln (\frac{25}{4} + \frac{- 25}{2} + 9)$

Step 1.4.1.2.1.7

Move the negative in front of the fraction.

$\ln (\frac{25}{4} - \frac{25}{2} + 9)$

Step 1.4.1.2.2

Find the common denominator.

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Step 1.4.1.2.2.1

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$\ln (\frac{25}{4} - (\frac{25}{2} \cdot \frac{2}{2}) + 9)$

Step 1.4.1.2.2.2

Multiply $\frac{25}{2}$ by $\frac{2}{2}$ .

$\ln (\frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + 9)$

Step 1.4.1.2.2.3

Write $9$ as a fraction with denominator $1$ .

$\ln (\frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{9}{1})$

Step 1.4.1.2.2.4

Multiply $\frac{9}{1}$ by $\frac{4}{4}$ .

$\ln (\frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{9}{1} \cdot \frac{4}{4})$

Step 1.4.1.2.2.5

Multiply $\frac{9}{1}$ by $\frac{4}{4}$ .

$\ln (\frac{25}{4} - \frac{25 \cdot 2}{2 \cdot 2} + \frac{9 \cdot 4}{4})$

Step 1.4.1.2.2.6

Multiply $2$ by $2$ .

$\ln (\frac{25}{4} - \frac{25 \cdot 2}{4} + \frac{9 \cdot 4}{4})$

Step 1.4.1.2.3

Combine the numerators over the common denominator.

$\ln (\frac{25 - 25 \cdot 2 + 9 \cdot 4}{4})$

Step 1.4.1.2.4

Simplify each term.

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Step 1.4.1.2.4.1

Multiply $- 25$ by $2$ .

$\ln (\frac{25 - 50 + 9 \cdot 4}{4})$

Step 1.4.1.2.4.2

Multiply $9$ by $4$ .

$\ln (\frac{25 - 50 + 36}{4})$

Step 1.4.1.2.5

Simplify by adding and subtracting.

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Step 1.4.1.2.5.1

Subtract $50$ from $25$ .

$\ln (\frac{- 25 + 36}{4})$

Step 1.4.1.2.5.2

Add $- 25$ and $36$ .

$\ln (\frac{11}{4})$

Step 1.4.2

List all of the points.

$(- \frac{5}{2}, \ln (\frac{11}{4}))$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $x = - 3$ .

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Step 2.1.1

Substitute $- 3$ for $x$ .

$\ln ({(- 3)}^{2} + 5 (- 3) + 9)$

Step 2.1.2

Simplify.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

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

$\ln (9 + 5 (- 3) + 9)$

Step 2.1.2.1.2

Multiply $5$ by $- 3$ .

$\ln (9 - 15 + 9)$

Step 2.1.2.2

Simplify by adding and subtracting.

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Step 2.1.2.2.1

Subtract $15$ from $9$ .

$\ln (- 6 + 9)$

Step 2.1.2.2.2

Add $- 6$ and $9$ .

$\ln (3)$

Step 2.2

Evaluate at $x = 1$ .

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Step 2.2.1

Substitute $1$ for $x$ .

$\ln ({(1)}^{2} + 5 (1) + 9)$

Step 2.2.2

Simplify.

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Step 2.2.2.1

Simplify each term.

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Step 2.2.2.1.1

One to any power is one.

$\ln (1 + 5 (1) + 9)$

Step 2.2.2.1.2

Multiply $5$ by $1$ .

$\ln (1 + 5 + 9)$

Step 2.2.2.2

Simplify by adding numbers.

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Step 2.2.2.2.1

Add $1$ and $5$ .

$\ln (6 + 9)$

Step 2.2.2.2.2

Add $6$ and $9$ .

$\ln (15)$

Step 2.3

List all of the points.

$(- 3, \ln (3)), (1, \ln (15))$

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.

Absolute Maximum: $(1, \ln (15))$

Absolute Minimum: $(- \frac{5}{2}, \ln (\frac{11}{4}))$

Step 4