Calculus Examples

$f (x) = \frac{x}{\ln (x)}$

Step 1

Find the second derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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 $f (x) = x$ and $g (x) = \ln (x)$ .

$f' (x) = \frac{\ln (x) \frac{d}{d x} (x) - x \frac{d}{d x} \ln (x)}{\ln^{2} (x)}$

Step 1.1.2

Differentiate using the Power Rule.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

$f' (x) = \frac{\ln (x) - x \frac{d}{d x} \ln (x)}{\ln^{2} (x)}$

Step 1.1.3

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

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

Step 1.1.4

Simplify terms.

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Step 1.1.4.1

Combine $\frac{1}{x}$ and $x$ .

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

Step 1.1.4.2

Cancel the common factor of $x$ .

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Step 1.1.4.2.1

Cancel the common factor.

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

Step 1.1.4.2.2

Rewrite the expression.

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

Step 1.1.4.3

Multiply $- 1$ by $1$ .

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

Step 1.2

Find the second derivative.

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Step 1.2.1

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

Step 1.2.2

Differentiate using the Sum Rule.

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Step 1.2.2.1

Multiply the exponents in ${(\ln^{2} (x))}^{2}$ .

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Step 1.2.2.1.1

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

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

Step 1.2.2.1.2

Multiply $2$ by $2$ .

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

Step 1.2.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Differentiate using the Constant Rule.

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Step 1.2.4.1

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

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

Step 1.2.4.2

Combine fractions.

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Step 1.2.4.2.1

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

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

Step 1.2.4.2.2

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

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

Step 1.2.5

Multiply $\frac{\frac{\ln^{2} (x)}{x} - (\ln (x) - 1) \frac{d}{d x} [\ln^{2} (x)]}{\ln^{4} (x)}$ by $\frac{x}{x}$ .

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

Step 1.2.6

Simplify terms.

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Step 1.2.6.1

Combine.

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

Step 1.2.6.2

Apply the distributive property.

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

Step 1.2.6.3

Cancel the common factor of $x$ .

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Step 1.2.6.3.1

Cancel the common factor.

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

Step 1.2.6.3.2

Rewrite the expression.

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

Step 1.2.7

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 $g (x) = \ln (x)$ .

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Step 1.2.7.1

To apply the Chain Rule, set $u$ as $\ln (x)$ .

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

Step 1.2.7.2

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

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

Step 1.2.7.3

Replace all occurrences of $u$ with $\ln (x)$ .

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

Step 1.2.8

Multiply $2$ by $- 1$ .

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

Step 1.2.9

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

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

Step 1.2.10

Simplify terms.

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Step 1.2.10.1

Combine $\ln (x)$ and $\frac{1}{x}$ .

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

Step 1.2.10.2

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

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

Step 1.2.10.3

Simplify the expression.

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Step 1.2.10.3.1

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

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

Step 1.2.10.3.2

Move the negative in front of the fraction.

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

Step 1.2.10.4

Combine $x$ and $\frac{2 \ln (x)}{x}$ .

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

Step 1.2.10.5

Cancel the common factor of $x$ .

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Step 1.2.10.5.1

Cancel the common factor.

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

Step 1.2.10.5.2

Divide $2 \ln (x)$ by $1$ .

$f'' (x) = \frac{\ln^{2} (x) - 2 \ln (x) (\ln (x) - 1)}{x \ln^{4} (x)}$

Step 1.2.10.6

Multiply $2$ by $- 1$ .

$f'' (x) = \frac{\ln^{2} (x) - 2 \ln (x) (\ln (x) - 1)}{x \ln^{4} (x)}$

Step 1.2.11

Simplify.

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Step 1.2.11.1

Apply the distributive property.

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

Step 1.2.11.2

Simplify each term.

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Step 1.2.11.2.1

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

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

Step 1.2.11.2.2

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

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

Step 1.2.11.2.3

Multiply $- \ln (x^{2}) \cdot - 1$ .

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Step 1.2.11.2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 1.2.11.2.3.2

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

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

Step 1.2.11.3

Reorder terms.

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

Step 1.3

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

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

Step 2

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

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Step 2.1

Set the second derivative equal to $0$ .

$\frac{\ln^{2} (x) - \ln (x) \ln (x^{2}) + \ln (x^{2})}{x \ln^{4} (x)} = 0$

Step 2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 7.38905639$

Step 3

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

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Step 3.1

Substitute $7.38905639$ in $f (x) = \frac{x}{\ln (x)}$ to find the value of $y$ .

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Step 3.1.1

Replace the variable $x$ with $7.38905639$ in the expression.

$f (7.38905639) = \frac{7.38905639}{\ln (7.38905639)}$

Step 3.1.2

Simplify the result.

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Step 3.1.2.1

Replace $e$ with an approximation.

$f (7.38905639) = \frac{7.38905639}{\log_{2.71828182} (7.38905639)}$

Step 3.1.2.2

Log base $2.71828182$ of $7.38905639$ is approximately $2.00000004$ .

$f (7.38905639) = \frac{7.38905639}{2.00000004}$

Step 3.1.2.3

Divide $7.38905639$ by $2.00000004$ .

$f (7.38905639) = 3.69452812$

Step 3.1.2.4

The final answer is $3.69452812$ .

$3.69452812$

Step 3.2

The point found by substituting $7.38905639$ in $f (x) = \frac{x}{\ln (x)}$ is $(7.38905639, 3.69452812)$ . This point can be an inflection point.

$(7.38905639, 3.69452812)$

Step 4

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

$(- \infty, 7.38905639) \cup (7.38905639, \infty)$

Step 5

Substitute a value from the interval $(- \infty, 7.38905639)$ into the second derivative to determine if it is increasing or decreasing.

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Step 5.1

Replace the variable $x$ with $7.28905639$ in the expression.

$f'' (7.28905639) = \frac{\ln^{2} (7.28905639) - \ln (7.28905639) \ln ({(7.28905639)}^{2}) + \ln ({(7.28905639)}^{2})}{(7.28905639) \ln^{4} (7.28905639)}$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify the numerator.

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Step 5.2.1.1

Raise $7.28905639$ to the power of $2$ .

$f'' (7.28905639) = \frac{\ln^{2} (7.28905639) - \ln (7.28905639) \ln (53.13034315) + \ln ({7.28905639}^{2})}{7.28905639 \ln^{4} (7.28905639)}$

Step 5.2.1.2

Raise $7.28905639$ to the power of $2$ .

$f'' (7.28905639) = \frac{\ln^{2} (7.28905639) - \ln (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{7.28905639 \ln^{4} (7.28905639)}$

Step 5.2.2

Multiply $7.28905639$ by $\ln^{4} (7.28905639)$ .

$f'' (7.28905639) = \frac{\ln^{2} (7.28905639) - \ln (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.3

Replace $e$ with an approximation.

$f'' (7.28905639) = \frac{{\log_{2.71828182}}^{2} (7.28905639) - \ln (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.4

Log base $2.71828182$ of $7.28905639$ is approximately $1.98637409$ .

$f'' (7.28905639) = \frac{{1.98637409}^{2} - \ln (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.5

Raise $1.98637409$ to the power of $2$ .

$f'' (7.28905639) = \frac{3.94568206 - \ln (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.6

Replace $e$ with an approximation.

$f'' (7.28905639) = \frac{3.94568206 - \log_{2.71828182} (7.28905639) \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.7

Log base $2.71828182$ of $7.28905639$ is approximately $1.98637409$ .

$f'' (7.28905639) = \frac{3.94568206 - 1 \cdot (1.98637409 \ln (53.13034315)) + \ln (53.13034315)}{113.47899623}$

Step 5.2.8

Multiply $- 1$ by $1.98637409$ .

$f'' (7.28905639) = \frac{3.94568206 - 1.98637409 \ln (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.9

Replace $e$ with an approximation.

$f'' (7.28905639) = \frac{3.94568206 - 1.98637409 \log_{2.71828182} (53.13034315) + \ln (53.13034315)}{113.47899623}$

Step 5.2.10

Log base $2.71828182$ of $53.13034315$ is approximately $3.97274819$ .

$f'' (7.28905639) = \frac{3.94568206 - 1.98637409 \cdot 3.97274819 + \ln (53.13034315)}{113.47899623}$

Step 5.2.11

Multiply $- 1.98637409$ by $3.97274819$ .

$f'' (7.28905639) = \frac{3.94568206 - 7.89136412 + \ln (53.13034315)}{113.47899623}$

Step 5.2.12

Subtract $7.89136412$ from $3.94568206$ .

$f'' (7.28905639) = \frac{- 3.94568206 + \ln (53.13034315)}{113.47899623}$

Step 5.2.13

Replace $e$ with an approximation.

$f'' (7.28905639) = \frac{- 3.94568206 + \log_{2.71828182} (53.13034315)}{113.47899623}$

Step 5.2.14

Log base $2.71828182$ of $53.13034315$ is approximately $3.97274819$ .

$f'' (7.28905639) = \frac{- 3.94568206 + 3.97274819}{113.47899623}$

Step 5.2.15

Add $- 3.94568206$ and $3.97274819$ .

$f'' (7.28905639) = \frac{0.02706613}{113.47899623}$

Step 5.2.16

Divide $0.02706613$ by $113.47899623$ .

$f'' (7.28905639) = 0.00023851$

Step 5.2.17

The final answer is $0.00023851$ .

$0.00023851$

Step 5.3

At $7.28905639$ , the second derivative is $0.00023851$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 7.38905639)$ .

Increasing on $(- \infty, 7.38905639)$ since $f'' (x) > 0$

Step 6

Substitute a value from the interval $(7.38905639, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $7.48905639$ in the expression.

$f'' (7.48905639) = \frac{\ln^{2} (7.48905639) - \ln (7.48905639) \ln ({(7.48905639)}^{2}) + \ln ({(7.48905639)}^{2})}{(7.48905639) \ln^{4} (7.48905639)}$

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 $7.48905639$ to the power of $2$ .

$f'' (7.48905639) = \frac{\ln^{2} (7.48905639) - \ln (7.48905639) \ln (56.0859657) + \ln ({7.48905639}^{2})}{7.48905639 \ln^{4} (7.48905639)}$

Step 6.2.1.2

Raise $7.48905639$ to the power of $2$ .

$f'' (7.48905639) = \frac{\ln^{2} (7.48905639) - \ln (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{7.48905639 \ln^{4} (7.48905639)}$

Step 6.2.2

Multiply $7.48905639$ by $\ln^{4} (7.48905639)$ .

$f'' (7.48905639) = \frac{\ln^{2} (7.48905639) - \ln (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.3

Replace $e$ with an approximation.

$f'' (7.48905639) = \frac{{\log_{2.71828182}}^{2} (7.48905639) - \ln (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.4

Log base $2.71828182$ of $7.48905639$ is approximately $2.0134428$ .

$f'' (7.48905639) = \frac{{2.0134428}^{2} - \ln (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.5

Raise $2.0134428$ to the power of $2$ .

$f'' (7.48905639) = \frac{4.05395194 - \ln (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.6

Replace $e$ with an approximation.

$f'' (7.48905639) = \frac{4.05395194 - \log_{2.71828182} (7.48905639) \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.7

Log base $2.71828182$ of $7.48905639$ is approximately $2.0134428$ .

$f'' (7.48905639) = \frac{4.05395194 - 1 \cdot (2.0134428 \ln (56.0859657)) + \ln (56.0859657)}{123.07909456}$

Step 6.2.8

Multiply $- 1$ by $2.0134428$ .

$f'' (7.48905639) = \frac{4.05395194 - 2.0134428 \ln (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.9

Replace $e$ with an approximation.

$f'' (7.48905639) = \frac{4.05395194 - 2.0134428 \log_{2.71828182} (56.0859657) + \ln (56.0859657)}{123.07909456}$

Step 6.2.10

Log base $2.71828182$ of $56.0859657$ is approximately $4.02688561$ .

$f'' (7.48905639) = \frac{4.05395194 - 2.0134428 \cdot 4.02688561 + \ln (56.0859657)}{123.07909456}$

Step 6.2.11

Multiply $- 2.0134428$ by $4.02688561$ .

$f'' (7.48905639) = \frac{4.05395194 - 8.10790388 + \ln (56.0859657)}{123.07909456}$

Step 6.2.12

Subtract $8.10790388$ from $4.05395194$ .

$f'' (7.48905639) = \frac{- 4.05395194 + \ln (56.0859657)}{123.07909456}$

Step 6.2.13

Replace $e$ with an approximation.

$f'' (7.48905639) = \frac{- 4.05395194 + \log_{2.71828182} (56.0859657)}{123.07909456}$

Step 6.2.14

Log base $2.71828182$ of $56.0859657$ is approximately $4.02688561$ .

$f'' (7.48905639) = \frac{- 4.05395194 + 4.02688561}{123.07909456}$

Step 6.2.15

Add $- 4.05395194$ and $4.02688561$ .

$f'' (7.48905639) = \frac{- 0.02706632}{123.07909456}$

Step 6.2.16

Divide $- 0.02706632$ by $123.07909456$ .

$f'' (7.48905639) = - 0.00021991$

Step 6.2.17

The final answer is $- 0.00021991$ .

$- 0.00021991$

Step 6.3

At $7.48905639$ , the second derivative is $- 0.00021991$ . Since this is negative, the second derivative is decreasing on the interval $(7.38905639, \infty)$

Decreasing on $(7.38905639, \infty)$ since $f'' (x) < 0$

Step 7

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 $(7.38905639, 3.69452812)$ .

$(7.38905639, 3.69452812)$

Step 8