Calculus Examples

$\frac{e^{x}}{8 + e^{x}}$

Step 1

Write $\frac{e^{x}}{8 + e^{x}}$ as a function.

$f (x) = \frac{e^{x}}{8 + e^{x}}$

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 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 $f (x) = e^{x}$ and $g (x) = 8 + e^{x}$ .

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

Step 2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.3

Differentiate.

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Step 2.1.3.1

By the Sum Rule, the derivative of $8 + e^{x}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [e^{x}]$ .

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

Step 2.1.3.2

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{(8 + e^{x}) e^{x} - e^{x} (0 + \frac{d}{d x} [e^{x}])}{{(8 + e^{x})}^{2}}$

Step 2.1.3.3

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

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

Step 2.1.4

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.5

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 2.1.5.1

Move $e^{x}$ .

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

Step 2.1.5.2

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

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

Step 2.1.5.3

Add $x$ and $x$ .

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

Step 2.1.6

Simplify.

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Step 2.1.6.1

Apply the distributive property.

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

Step 2.1.6.2

Simplify the numerator.

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Step 2.1.6.2.1

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 2.1.6.2.1.1

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

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

Step 2.1.6.2.1.2

Add $x$ and $x$ .

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

Step 2.1.6.2.2

Combine the opposite terms in $8 e^{x} + e^{2 x} - e^{2 x}$ .

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Step 2.1.6.2.2.1

Subtract $e^{2 x}$ from $e^{2 x}$ .

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

Step 2.1.6.2.2.2

Add $8 e^{x}$ and $0$ .

$f' (x) = \frac{8 e^{x}}{{(8 + e^{x})}^{2}}$

Step 2.2

Find the second derivative.

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Step 2.2.1

Since $8$ is constant with respect to $x$ , the derivative of $\frac{8 e^{x}}{{(8 + e^{x})}^{2}}$ with respect to $x$ is $8 \frac{d}{d x} [\frac{e^{x}}{{(8 + e^{x})}^{2}}]$ .

$8 \frac{d}{d x} [\frac{e^{x}}{{(8 + e^{x})}^{2}}]$

Step 2.2.2

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

Step 2.2.3

Multiply the exponents in ${({(8 + e^{x})}^{2})}^{2}$ .

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Step 2.2.3.1

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

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

Step 2.2.3.2

Multiply $2$ by $2$ .

$8 \frac{{(8 + e^{x})}^{2} \frac{d}{d x} [e^{x}] - e^{x} \frac{d}{d x} [{(8 + e^{x})}^{2}]}{{(8 + e^{x})}^{4}}$

Step 2.2.4

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - e^{x} \frac{d}{d x} [{(8 + e^{x})}^{2}]}{{(8 + e^{x})}^{4}}$

Step 2.2.5

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 $g (x) = 8 + e^{x}$ .

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Step 2.2.5.1

To apply the Chain Rule, set $u$ as $8 + e^{x}$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - e^{x} (\frac{d}{d u} [u^{2}] \frac{d}{d x} [8 + e^{x}])}{{(8 + e^{x})}^{4}}$

Step 2.2.5.2

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

$8 \frac{{(8 + e^{x})}^{2} e^{x} - e^{x} (2 u \frac{d}{d x} [8 + e^{x}])}{{(8 + e^{x})}^{4}}$

Step 2.2.5.3

Replace all occurrences of $u$ with $8 + e^{x}$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - e^{x} (2 (8 + e^{x}) \frac{d}{d x} [8 + e^{x}])}{{(8 + e^{x})}^{4}}$

Step 2.2.6

Differentiate.

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Step 2.2.6.1

Multiply $2$ by $- 1$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x} ((8 + e^{x}) \frac{d}{d x} [8 + e^{x}])}{{(8 + e^{x})}^{4}}$

Step 2.2.6.2

By the Sum Rule, the derivative of $8 + e^{x}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [e^{x}]$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x} ((8 + e^{x}) (\frac{d}{d x} [8] + \frac{d}{d x} [e^{x}]))}{{(8 + e^{x})}^{4}}$

Step 2.2.6.3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x} ((8 + e^{x}) (0 + \frac{d}{d x} [e^{x}]))}{{(8 + e^{x})}^{4}}$

Step 2.2.6.4

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

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x} ((8 + e^{x}) \frac{d}{d x} [e^{x}])}{{(8 + e^{x})}^{4}}$

Step 2.2.7

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x} ((8 + e^{x}) e^{x})}{{(8 + e^{x})}^{4}}$

Step 2.2.8

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

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{x + x} (8 + e^{x})}{{(8 + e^{x})}^{4}}$

Step 2.2.9

Add $x$ and $x$ .

$8 \frac{{(8 + e^{x})}^{2} e^{x} - 2 e^{2 x} (8 + e^{x})}{{(8 + e^{x})}^{4}}$

Step 2.2.10

Factor $8 + e^{x}$ out of ${(8 + e^{x})}^{2} e^{x} - 2 e^{2 x} (8 + e^{x})$ .

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Step 2.2.10.1

Factor $8 + e^{x}$ out of ${(8 + e^{x})}^{2} e^{x}$ .

$8 \frac{(8 + e^{x}) ((8 + e^{x}) e^{x}) - 2 e^{2 x} (8 + e^{x})}{{(8 + e^{x})}^{4}}$

Step 2.2.10.2

Factor $8 + e^{x}$ out of $- 2 e^{2 x} (8 + e^{x})$ .

$8 \frac{(8 + e^{x}) ((8 + e^{x}) e^{x}) + (8 + e^{x}) (- 2 e^{2 x})}{{(8 + e^{x})}^{4}}$

Step 2.2.10.3

Factor $8 + e^{x}$ out of $(8 + e^{x}) ((8 + e^{x}) e^{x}) + (8 + e^{x}) (- 2 e^{2 x})$ .

$8 \frac{(8 + e^{x}) ((8 + e^{x}) e^{x} - 2 e^{2 x})}{{(8 + e^{x})}^{4}}$

Step 2.2.11

Cancel the common factors.

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Step 2.2.11.1

Factor $8 + e^{x}$ out of ${(8 + e^{x})}^{4}$ .

$8 \frac{(8 + e^{x}) ((8 + e^{x}) e^{x} - 2 e^{2 x})}{(8 + e^{x}) {(8 + e^{x})}^{3}}$

Step 2.2.11.2

Cancel the common factor.

$8 \frac{(8 + e^{x}) ((8 + e^{x}) e^{x} - 2 e^{2 x})}{(8 + e^{x}) {(8 + e^{x})}^{3}}$

Step 2.2.11.3

Rewrite the expression.

$8 \frac{(8 + e^{x}) e^{x} - 2 e^{2 x}}{{(8 + e^{x})}^{3}}$

Step 2.2.12

Combine $8$ and $\frac{(8 + e^{x}) e^{x} - 2 e^{2 x}}{{(8 + e^{x})}^{3}}$ .

$\frac{8 ((8 + e^{x}) e^{x} - 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13

Simplify.

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Step 2.2.13.1

Apply the distributive property.

$\frac{8 (8 e^{x} + e^{x} e^{x} - 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.2

Apply the distributive property.

$\frac{8 (8 e^{x}) + 8 (e^{x} e^{x}) + 8 (- 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.3

Simplify the numerator.

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Step 2.2.13.3.1

Simplify each term.

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Step 2.2.13.3.1.1

Multiply $8$ by $8$ .

$\frac{64 e^{x} + 8 (e^{x} e^{x}) + 8 (- 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.3.1.2

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 2.2.13.3.1.2.1

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

$\frac{64 e^{x} + 8 e^{x + x} + 8 (- 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.3.1.2.2

Add $x$ and $x$ .

$\frac{64 e^{x} + 8 e^{2 x} + 8 (- 2 e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.3.1.3

Multiply $- 2$ by $8$ .

$\frac{64 e^{x} + 8 e^{2 x} - 16 e^{2 x}}{{(8 + e^{x})}^{3}}$

Step 2.2.13.3.2

Subtract $16 e^{2 x}$ from $8 e^{2 x}$ .

$\frac{64 e^{x} - 8 e^{2 x}}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4

Simplify the numerator.

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Step 2.2.13.4.1

Factor $8$ out of $64 e^{x} - 8 e^{2 x}$ .

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Step 2.2.13.4.1.1

Factor $8$ out of $64 e^{x}$ .

$\frac{8 (8 e^{x}) - 8 e^{2 x}}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.1.2

Factor $8$ out of $- 8 e^{2 x}$ .

$\frac{8 (8 e^{x}) + 8 (- e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.1.3

Factor $8$ out of $8 (8 e^{x}) + 8 (- e^{2 x})$ .

$\frac{8 (8 e^{x} - e^{2 x})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.2

Rewrite $e^{2 x}$ as ${(e^{x})}^{2}$ .

$\frac{8 (8 e^{x} - {(e^{x})}^{2})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.3

Let $u = e^{x}$ . Substitute $u$ for all occurrences of $e^{x}$ .

$\frac{8 (8 u - u^{2})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.4

Factor $u$ out of $8 u - u^{2}$ .

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Step 2.2.13.4.4.1

Factor $u$ out of $8 u$ .

$\frac{8 (u \cdot 8 - u^{2})}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.4.2

Factor $u$ out of $- u^{2}$ .

$\frac{8 (u \cdot 8 + u (- u))}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.4.3

Factor $u$ out of $u \cdot 8 + u (- u)$ .

$\frac{8 (u (8 - u))}{{(8 + e^{x})}^{3}}$

Step 2.2.13.4.5

Replace all occurrences of $u$ with $e^{x}$ .

$f'' (x) = \frac{8 e^{x} (8 - e^{x})}{{(8 + e^{x})}^{3}}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{8 e^{x} (8 - e^{x})}{{(8 + e^{x})}^{3}}$ .

$\frac{8 e^{x} (8 - e^{x})}{{(8 + e^{x})}^{3}}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{8 e^{x} (8 - e^{x})}{{(8 + e^{x})}^{3}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$\frac{8 e^{x} (8 - e^{x})}{{(8 + e^{x})}^{3}} = 0$

Step 3.2

Set the numerator equal to zero.

$8 e^{x} (8 - e^{x}) = 0$

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 $0$ .

$e^{x} = 0$

$8 - e^{x} = 0$

Step 3.3.2

Set $e^{x}$ equal to $0$ and solve for $x$ .

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Step 3.3.2.1

Set $e^{x}$ equal to $0$ .

$e^{x} = 0$

Step 3.3.2.2

Solve $e^{x} = 0$ for $x$ .

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Step 3.3.2.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x}) = \ln (0)$

Step 3.3.2.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 3.3.2.2.3

There is no solution for $e^{x} = 0$

No solution

Step 3.3.3

Set $8 - e^{x}$ equal to $0$ and solve for $x$ .

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Step 3.3.3.1

Set $8 - e^{x}$ equal to $0$ .

$8 - e^{x} = 0$

Step 3.3.3.2

Solve $8 - e^{x} = 0$ for $x$ .

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Step 3.3.3.2.1

Subtract $8$ from both sides of the equation.

$- e^{x} = - 8$

Step 3.3.3.2.2

Divide each term in $- e^{x} = - 8$ by $- 1$ and simplify.

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Step 3.3.3.2.2.1

Divide each term in $- e^{x} = - 8$ by $- 1$ .

$\frac{- e^{x}}{- 1} = \frac{- 8}{- 1}$

Step 3.3.3.2.2.2

Simplify the left side.

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Step 3.3.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{e^{x}}{1} = \frac{- 8}{- 1}$

Step 3.3.3.2.2.2.2

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

$e^{x} = \frac{- 8}{- 1}$

Step 3.3.3.2.2.3

Simplify the right side.

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Step 3.3.3.2.2.3.1

Divide $- 8$ by $- 1$ .

$e^{x} = 8$

Step 3.3.3.2.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{x}) = \ln (8)$

Step 3.3.3.2.4

Expand the left side.

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Step 3.3.3.2.4.1

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$x \ln (e) = \ln (8)$

Step 3.3.3.2.4.2

The natural logarithm of $e$ is $1$ .

$x \cdot 1 = \ln (8)$

Step 3.3.3.2.4.3

Multiply $x$ by $1$ .

$x = \ln (8)$

Step 3.3.4

The final solution is all the values that make $8 e^{x} (8 - e^{x}) = 0$ true.

$x = \ln (8)$

Step 4

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

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Step 4.1

Substitute $\ln (8)$ in $f (x) = \frac{e^{x}}{8 + e^{x}}$ to find the value of $y$ .

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Step 4.1.1

Replace the variable $x$ with $\ln (8)$ in the expression.

$f (\ln (8)) = \frac{e^{\ln (8)}}{8 + e^{\ln (8)}}$

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Exponentiation and log are inverse functions.

$f (\ln (8)) = \frac{8}{8 + e^{\ln (8)}}$

Step 4.1.2.2

Simplify the denominator.

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Step 4.1.2.2.1

Exponentiation and log are inverse functions.

$f (\ln (8)) = \frac{8}{8 + 8}$

Step 4.1.2.2.2

Add $8$ and $8$ .

$f (\ln (8)) = \frac{8}{16}$

Step 4.1.2.3

Cancel the common factor of $8$ and $16$ .

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Step 4.1.2.3.1

Factor $8$ out of $8$ .

$f (\ln (8)) = \frac{8 (1)}{16}$

Step 4.1.2.3.2

Cancel the common factors.

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Step 4.1.2.3.2.1

Factor $8$ out of $16$ .

$f (\ln (8)) = \frac{8 \cdot 1}{8 \cdot 2}$

Step 4.1.2.3.2.2

Cancel the common factor.

$f (\ln (8)) = \frac{8 \cdot 1}{8 \cdot 2}$

Step 4.1.2.3.2.3

Rewrite the expression.

$f (\ln (8)) = \frac{1}{2}$

Step 4.1.2.4

The final answer is $\frac{1}{2}$ .

$\frac{1}{2}$

Step 4.2

The point found by substituting $\ln (8)$ in $f (x) = \frac{e^{x}}{8 + e^{x}}$ is $(\ln (8), \frac{1}{2})$ . This point can be an inflection point.

$(\ln (8), \frac{1}{2})$

Step 5

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

$(- \infty, \ln (8)) \cup (\ln (8), \infty)$

Step 6

Substitute a value from the interval $(- \infty, \ln (8))$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $1.97944154$ in the expression.

$f'' (1.97944154) = \frac{8 e^{1.97944154} (8 - e^{1.97944154})}{{(8 + e^{1.97944154})}^{3}}$

Step 6.2

The final answer is $\frac{8 e^{1.97944154} (8 - e^{1.97944154})}{{(8 + e^{1.97944154})}^{3}}$ .

$\frac{8 e^{1.97944154} (8 - e^{1.97944154})}{{(8 + e^{1.97944154})}^{3}}$

Step 6.3

At $1.97944154$ , the second derivative is $0.01245842$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, \ln (8))$ .

Increasing on $(- \infty, \ln (8))$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(\ln (8), \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $2.17944154$ in the expression.

$f'' (2.17944154) = \frac{8 e^{2.17944154} (8 - e^{2.17944154})}{{(8 + e^{2.17944154})}^{3}}$

Step 7.2

The final answer is $\frac{8 e^{2.17944154} (8 - e^{2.17944154})}{{(8 + e^{2.17944154})}^{3}}$ .

$\frac{8 e^{2.17944154} (8 - e^{2.17944154})}{{(8 + e^{2.17944154})}^{3}}$

Step 7.3

At $2.17944154$ , the second derivative is $- 0.01245842$ . Since this is negative, the second derivative is decreasing on the interval $(\ln (8), \infty)$

Decreasing on $(\ln (8), \infty)$ since $f'' (x) < 0$

Step 8

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 $(\ln (8), \frac{1}{2})$ .

$(\ln (8), \frac{1}{2})$

Step 9