Calculus Examples

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

Step 1

Write $\log_{5} (1 + x^{2})$ as a function.

$f (x) = \log_{5} (1 + x^{2})$

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 chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \log_{5} (x)$ and $g (x) = 1 + x^{2}$ .

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Step 2.1.1.1

To apply the Chain Rule, set $u$ as $1 + x^{2}$ .

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

Step 2.1.1.2

The derivative of $\log_{5} (u)$ with respect to $u$ is $\frac{1}{u \ln (5)}$ .

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

Step 2.1.1.3

Replace all occurrences of $u$ with $1 + x^{2}$ .

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

Step 2.1.2

Differentiate.

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Add $0$ and $\frac{d}{d x} [x^{2}]$ .

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

Step 2.1.2.4

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

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

Step 2.1.2.5

Combine fractions.

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Step 2.1.2.5.1

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

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

Step 2.1.2.5.2

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

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

Step 2.1.3

Simplify.

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Step 2.1.3.1

Apply the distributive property.

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

Step 2.1.3.2

Multiply $\ln (5)$ by $1$ .

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

Step 2.1.3.3

Reorder terms.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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 $g (x) = x^{2} \ln (5) + \ln (5)$ .

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

Step 2.2.3

Differentiate.

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.3.4

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

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

Step 2.2.3.5

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

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

Step 2.2.3.6

Move $2$ to the left of $\ln (5)$ .

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

Step 2.2.3.7

Since $\ln (5)$ is constant with respect to $x$ , the derivative of $\ln (5)$ with respect to $x$ is $0$ .

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

Step 2.2.3.8

Simplify the expression.

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Step 2.2.3.8.1

Add $2 \ln (5) x$ and $0$ .

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

Step 2.2.3.8.2

Multiply $2$ by $- 1$ .

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Add $1$ and $1$ .

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

Step 2.2.8

Subtract $2 x^{2} \ln (5)$ from $x^{2} \ln (5)$ .

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

Step 2.2.9

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

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

Step 2.2.10

Simplify.

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Step 2.2.10.1

Apply the distributive property.

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

Step 2.2.10.2

Simplify the numerator.

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Step 2.2.10.2.1

Simplify each term.

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Step 2.2.10.2.1.1

Multiply $2 (- x^{2} \ln (5))$ .

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Step 2.2.10.2.1.1.1

Multiply $- 1$ by $2$ .

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

Step 2.2.10.2.1.1.2

Simplify $- 2 \ln (5)$ by moving $2$ inside the logarithm.

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

Step 2.2.10.2.1.2

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

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

Step 2.2.10.2.1.3

Simplify $2 \ln (5)$ by moving $2$ inside the logarithm.

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

Step 2.2.10.2.1.4

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

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

Step 2.2.10.2.2

Reorder factors in $x^{2} (- \ln (25)) + \ln (25)$ .

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

Step 2.2.10.3

Factor $- 1$ out of $- x^{2} \ln (25)$ .

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

Step 2.2.10.4

Factor $- 1$ out of $\ln (25)$ .

$\frac{- (x^{2} \ln (25)) - 1 (- \ln (25))}{{(x^{2} \ln (5) + \ln (5))}^{2}}$

Step 2.2.10.5

Factor $- 1$ out of $- (x^{2} \ln (25)) - 1 (- \ln (25))$ .

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

Step 2.2.10.6

Rewrite $- (x^{2} \ln (25) - \ln (25))$ as $- 1 (x^{2} \ln (25) - \ln (25))$ .

$\frac{- 1 (x^{2} \ln (25) - \ln (25))}{{(x^{2} \ln (5) + \ln (5))}^{2}}$

Step 2.2.10.7

Move the negative in front of the fraction.

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

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{x^{2} \ln (25) - \ln (25)}{{(x^{2} \ln (5) + \ln (5))}^{2}}$ .

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

Step 3

Set the second derivative equal to $0$ then solve the equation $- \frac{x^{2} \ln (25) - \ln (25)}{{(x^{2} \ln (5) + \ln (5))}^{2}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$- \frac{x^{2} \ln (25) - \ln (25)}{{(x^{2} \ln (5) + \ln (5))}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$x^{2} \ln (25) - \ln (25) = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

Add $\ln (25)$ to both sides of the equation.

$x^{2} \ln (25) = \ln (25)$

Step 3.3.2

Divide each term in $x^{2} \ln (25) = \ln (25)$ by $\ln (25)$ and simplify.

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Step 3.3.2.1

Divide each term in $x^{2} \ln (25) = \ln (25)$ by $\ln (25)$ .

$\frac{x^{2} \ln (25)}{\ln (25)} = \frac{\ln (25)}{\ln (25)}$

Step 3.3.2.2

Simplify the left side.

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Step 3.3.2.2.1

Cancel the common factor of $\ln (25)$ .

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Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{x^{2} \ln (25)}{\ln (25)} = \frac{\ln (25)}{\ln (25)}$

Step 3.3.2.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{\ln (25)}{\ln (25)}$

Step 3.3.2.3

Simplify the right side.

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Step 3.3.2.3.1

Cancel the common factor of $\ln (25)$ .

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Step 3.3.2.3.1.1

Cancel the common factor.

$x^{2} = \frac{\ln (25)}{\ln (25)}$

Step 3.3.2.3.1.2

Rewrite the expression.

$x^{2} = 1$

Step 3.3.3

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

$x = \pm \sqrt{1}$

Step 3.3.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 3.3.5

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

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Step 3.3.5.1

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

$x = 1$

Step 3.3.5.2

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

$x = - 1$

Step 3.3.5.3

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

$x = 1, - 1$

Step 4

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

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Step 4.1

Substitute $1$ in $f (x) = \log_{5} (1 + x^{2})$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \log_{5} (1 + {(1)}^{2})$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

One to any power is one.

$f (1) = \log_{5} (1 + 1)$

Step 4.1.2.2

Add $1$ and $1$ .

$f (1) = \log_{5} (2)$

Step 4.1.2.3

The final answer is $\log_{5} (2)$ .

$\log_{5} (2)$

Step 4.2

The point found by substituting $1$ in $f (x) = \log_{5} (1 + x^{2})$ is $(1, \log_{5} (2))$ . This point can be an inflection point.

$(1, \log_{5} (2))$

Step 4.3

Substitute $- 1$ in $f (x) = \log_{5} (1 + x^{2})$ to find the value of $y$ .

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Step 4.3.1

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = \log_{5} (1 + {(- 1)}^{2})$

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

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

$f (- 1) = \log_{5} (1 + 1)$

Step 4.3.2.2

Add $1$ and $1$ .

$f (- 1) = \log_{5} (2)$

Step 4.3.2.3

The final answer is $\log_{5} (2)$ .

$\log_{5} (2)$

Step 4.4

The point found by substituting $- 1$ in $f (x) = \log_{5} (1 + x^{2})$ is $(- 1, \log_{5} (2))$ . This point can be an inflection point.

$(- 1, \log_{5} (2))$

Step 4.5

Determine the points that could be inflection points.

$(1, \log_{5} (2)), (- 1, \log_{5} (2))$

Step 5

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

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = - \frac{{(- 1.1)}^{2} \ln (25) - \ln (25)}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

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

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = - \frac{1.21 \ln (25) - \ln (25)}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

Step 6.2.1.2

Simplify $1.21 \ln (25)$ by moving $1.21$ inside the logarithm.

$f'' (- 1.1) = - \frac{\ln (25^{1.21}) - \ln (25)}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

Step 6.2.1.3

Raise $25$ to the power of $1.21$ .

$f'' (- 1.1) = - \frac{\ln (49.14817664) - \ln (25)}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

Step 6.2.1.4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$f'' (- 1.1) = - \frac{\ln (\frac{49.14817664}{25})}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

Step 6.2.1.5

Divide $49.14817664$ by $25$ .

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{{({(- 1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

Step 6.2.2

Simplify the denominator.

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Step 6.2.2.1

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{{(1.21 \ln (5) + \ln (5))}^{2}}$

Step 6.2.2.2

Simplify $1.21 \ln (5)$ by moving $1.21$ inside the logarithm.

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{{(\ln (5^{1.21}) + \ln (5))}^{2}}$

Step 6.2.2.3

Raise $5$ to the power of $1.21$ .

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{{(\ln (7.01057605) + \ln (5))}^{2}}$

Step 6.2.2.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{\ln^{2} (7.01057605 \cdot 5)}$

Step 6.2.2.5

Multiply $7.01057605$ by $5$ .

$f'' (- 1.1) = - \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$

Step 6.2.3

The final answer is $- \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$ .

$- \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$

Step 6.3

At $- 1.1$ , the second derivative is $- 0.05343065$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, - 1)$

Decreasing on $(- \infty, - 1)$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(- 1, 1)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

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

$f'' (0) = - \frac{{(0)}^{2} \ln (25) - \ln (25)}{{({(0)}^{2} \ln (5) + \ln (5))}^{2}}$

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

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

$f'' (0) = - \frac{0 \ln (25) - \ln (25)}{{(0^{2} \ln (5) + \ln (5))}^{2}}$

Step 7.2.1.2

Multiply $0$ by $\ln (25)$ .

$f'' (0) = - \frac{0 - \ln (25)}{{(0^{2} \ln (5) + \ln (5))}^{2}}$

Step 7.2.1.3

Subtract $\ln (25)$ from $0$ .

$f'' (0) = - \frac{- \ln (25)}{{(0^{2} \ln (5) + \ln (5))}^{2}}$

Step 7.2.2

Simplify the denominator.

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Step 7.2.2.1

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

$f'' (0) = - \frac{- \ln (25)}{{(0 \ln (5) + \ln (5))}^{2}}$

Step 7.2.2.2

Multiply $0$ by $\ln (5)$ .

$f'' (0) = - \frac{- \ln (25)}{{(0 + \ln (5))}^{2}}$

Step 7.2.2.3

Add $0$ and $\ln (5)$ .

$f'' (0) = - \frac{- \ln (25)}{\ln^{2} (5)}$

Step 7.2.3

Move the negative in front of the fraction.

$f'' (0) = \frac{\ln (25)}{\ln^{2} (5)}$

Step 7.2.4

The final answer is $\frac{\ln (25)}{\ln^{2} (5)}$ .

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

Step 7.3

At $0$ , the second derivative is $1.24266986$ . Since this is positive, the second derivative is increasing on the interval $(- 1, 1)$ .

Increasing on $(- 1, 1)$ since $f'' (x) > 0$

Step 8

Substitute a value from the interval $(1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $1.1$ in the expression.

$f'' (1.1) = - \frac{{(1.1)}^{2} \ln (25) - \ln (25)}{{({(1.1)}^{2} \ln (5) + \ln (5))}^{2}}$

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

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = - \frac{1.21 \ln (25) - \ln (25)}{{({1.1}^{2} \ln (5) + \ln (5))}^{2}}$

Step 8.2.1.2

Simplify $1.21 \ln (25)$ by moving $1.21$ inside the logarithm.

$f'' (1.1) = - \frac{\ln (25^{1.21}) - \ln (25)}{{({1.1}^{2} \ln (5) + \ln (5))}^{2}}$

Step 8.2.1.3

Raise $25$ to the power of $1.21$ .

$f'' (1.1) = - \frac{\ln (49.14817664) - \ln (25)}{{({1.1}^{2} \ln (5) + \ln (5))}^{2}}$

Step 8.2.1.4

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$f'' (1.1) = - \frac{\ln (\frac{49.14817664}{25})}{{({1.1}^{2} \ln (5) + \ln (5))}^{2}}$

Step 8.2.1.5

Divide $49.14817664$ by $25$ .

$f'' (1.1) = - \frac{\ln (1.96592706)}{{({1.1}^{2} \ln (5) + \ln (5))}^{2}}$

Step 8.2.2

Simplify the denominator.

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Step 8.2.2.1

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = - \frac{\ln (1.96592706)}{{(1.21 \ln (5) + \ln (5))}^{2}}$

Step 8.2.2.2

Simplify $1.21 \ln (5)$ by moving $1.21$ inside the logarithm.

$f'' (1.1) = - \frac{\ln (1.96592706)}{{(\ln (5^{1.21}) + \ln (5))}^{2}}$

Step 8.2.2.3

Raise $5$ to the power of $1.21$ .

$f'' (1.1) = - \frac{\ln (1.96592706)}{{(\ln (7.01057605) + \ln (5))}^{2}}$

Step 8.2.2.4

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$f'' (1.1) = - \frac{\ln (1.96592706)}{\ln^{2} (7.01057605 \cdot 5)}$

Step 8.2.2.5

Multiply $7.01057605$ by $5$ .

$f'' (1.1) = - \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$

Step 8.2.3

The final answer is $- \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$ .

$- \frac{\ln (1.96592706)}{\ln^{2} (35.05288028)}$

Step 8.3

At $1.1$ , the second derivative is $- 0.05343065$ . Since this is negative, the second derivative is decreasing on the interval $(1, \infty)$

Decreasing on $(1, \infty)$ 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 points in this case are $(- 1, \log_{5} (2)), (1, \log_{5} (2))$ .

$(- 1, \log_{5} (2)), (1, \log_{5} (2))$

Step 10