Calculus Examples

$y = 5 x^{2} \ln (\frac{x}{4})$

Step 1

Write $y = 5 x^{2} \ln (\frac{x}{4})$ as a function.

$f (x) = 5 x^{2} \ln (\frac{x}{4})$

Step 2

Find the second derivative.

Tap for more steps...

Step 2.1

Find the first derivative.

Tap for more steps...

Step 2.1.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2} \ln (\frac{x}{4})$ with respect to $x$ is $5 \frac{d}{d x} [x^{2} \ln (\frac{x}{4})]$ .

$f' (x) = 5 \frac{d}{d x} (x^{2} \ln (\frac{x}{4}))$

Step 2.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = \ln (\frac{x}{4})$ .

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

Step 2.1.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \frac{x}{4}$ .

Tap for more steps...

Step 2.1.3.1

To apply the Chain Rule, set $u$ as $\frac{x}{4}$ .

$f' (x) = 5 (x^{2} (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (\frac{x}{4})) + \ln (\frac{x}{4}) \frac{d}{d x} (x^{2}))$

Step 2.1.3.2

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

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

Step 2.1.3.3

Replace all occurrences of $u$ with $\frac{x}{4}$ .

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

Step 2.1.4

Multiply by the reciprocal of the fraction to divide by $\frac{x}{4}$ .

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

Step 2.1.5

Simplify terms.

Tap for more steps...

Step 2.1.5.1

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

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

Step 2.1.5.2

Combine $\frac{4}{x}$ and $x^{2}$ .

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

Step 2.1.5.3

Cancel the common factor of $x^{2}$ and $x$ .

Tap for more steps...

Step 2.1.5.3.1

Factor $x$ out of $4 x^{2}$ .

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

Step 2.1.5.3.2

Cancel the common factors.

Tap for more steps...

Step 2.1.5.3.2.1

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

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

Step 2.1.5.3.2.2

Factor $x$ out of $x^{1}$ .

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

Step 2.1.5.3.2.3

Cancel the common factor.

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

Step 2.1.5.3.2.4

Rewrite the expression.

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

Step 2.1.5.3.2.5

Divide $4 x$ by $1$ .

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

Step 2.1.6

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{x}{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x]$ .

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

Step 2.1.7

Simplify terms.

Tap for more steps...

Step 2.1.7.1

Combine $4$ and $\frac{1}{4}$ .

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

Step 2.1.7.2

Combine $\frac{4}{4}$ and $x$ .

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

Step 2.1.7.3

Cancel the common factor of $4$ .

Tap for more steps...

Step 2.1.7.3.1

Cancel the common factor.

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

Step 2.1.7.3.2

Divide $x$ by $1$ .

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

Step 2.1.8

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

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

Step 2.1.9

Multiply $x$ by $1$ .

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

Step 2.1.10

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

$f' (x) = 5 (x + \ln (\frac{x}{4}) (2 x))$

Step 2.1.11

Simplify.

Tap for more steps...

Step 2.1.11.1

Apply the distributive property.

$f' (x) = 5 x + 5 (\ln (\frac{x}{4}) (2 x))$

Step 2.1.11.2

Multiply $2$ by $5$ .

$f' (x) = 5 x + 10 \ln (\frac{x}{4}) x$

Step 2.1.11.3

Reorder terms.

$f' (x) = 10 x \ln (\frac{x}{4}) + 5 x$

Step 2.2

Find the second derivative.

Tap for more steps...

Step 2.2.1

By the Sum Rule, the derivative of $10 x \ln (\frac{x}{4}) + 5 x$ with respect to $x$ is $\frac{d}{d x} [10 x \ln (\frac{x}{4})] + \frac{d}{d x} [5 x]$ .

$f'' (x) = \frac{d}{d x} (10 x \ln (\frac{x}{4})) + \frac{d}{d x} (5 x)$

Step 2.2.2

Evaluate $\frac{d}{d x} [10 x \ln (\frac{x}{4})]$ .

Tap for more steps...

Step 2.2.2.1

Since $10$ is constant with respect to $x$ , the derivative of $10 x \ln (\frac{x}{4})$ with respect to $x$ is $10 \frac{d}{d x} [x \ln (\frac{x}{4})]$ .

$f'' (x) = 10 \frac{d}{d x} (x \ln (\frac{x}{4})) + \frac{d}{d x} (5 x)$

Step 2.2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (\frac{x}{4})$ .

$f'' (x) = 10 (x \frac{d}{d x} (\ln (\frac{x}{4})) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = \frac{x}{4}$ .

Tap for more steps...

Step 2.2.2.3.1

To apply the Chain Rule, set $u$ as $\frac{x}{4}$ .

$f'' (x) = 10 (x (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (\frac{x}{4})) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.3.2

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

$f'' (x) = 10 (x (\frac{1}{u} \cdot \frac{d}{d x} (\frac{x}{4})) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.3.3

Replace all occurrences of $u$ with $\frac{x}{4}$ .

$f'' (x) = 10 (x (\frac{1}{\frac{x}{4}} \cdot \frac{d}{d x} (\frac{x}{4})) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.4

Since $\frac{1}{4}$ is constant with respect to $x$ , the derivative of $\frac{x}{4}$ with respect to $x$ is $\frac{1}{4} \frac{d}{d x} [x]$ .

$f'' (x) = 10 (x (\frac{1}{\frac{x}{4}} \cdot (\frac{1}{4} \cdot \frac{d}{d x} (x))) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.5

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

$f'' (x) = 10 (x (\frac{1}{\frac{x}{4}} \cdot (\frac{1}{4} \cdot 1)) + \ln (\frac{x}{4}) \frac{d}{d x} (x)) + \frac{d}{d x} (5 x)$

Step 2.2.2.6

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

$f'' (x) = 10 (x (\frac{1}{\frac{x}{4}} \cdot (\frac{1}{4} \cdot 1)) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.7

Multiply by the reciprocal of the fraction to divide by $\frac{x}{4}$ .

$f'' (x) = 10 (x (1 (\frac{4}{x}) \cdot (\frac{1}{4} \cdot 1)) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.8

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

$f'' (x) = 10 (x (\frac{4}{x} \cdot (\frac{1}{4} \cdot 1)) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.9

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

$f'' (x) = 10 (x (\frac{4}{x} \cdot \frac{1}{4}) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.10

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

$f'' (x) = 10 (x (\frac{4}{x \cdot 4}) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.11

Move $4$ to the left of $x$ .

$f'' (x) = 10 (x (\frac{4}{4 \cdot x}) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.12

Cancel the common factor of $4$ .

Tap for more steps...

Step 2.2.2.12.1

Cancel the common factor.

$f'' (x) = 10 (x (\frac{4}{4 x}) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.12.2

Rewrite the expression.

$f'' (x) = 10 (x (\frac{1}{x}) + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.13

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

$f'' (x) = 10 (\frac{x}{x} + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.14

Cancel the common factor of $x$ .

Tap for more steps...

Step 2.2.2.14.1

Cancel the common factor.

$f'' (x) = 10 (\frac{x}{x} + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.14.2

Rewrite the expression.

$f'' (x) = 10 (1 + \ln (\frac{x}{4}) \cdot 1) + \frac{d}{d x} (5 x)$

Step 2.2.2.15

Multiply $\ln (\frac{x}{4})$ by $1$ .

$f'' (x) = 10 (1 + \ln (\frac{x}{4})) + \frac{d}{d x} (5 x)$

Step 2.2.3

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

Tap for more steps...

Step 2.2.3.1

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

$f'' (x) = 10 (1 + \ln (\frac{x}{4})) + 5 \frac{d}{d x} (x)$

Step 2.2.3.2

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

$f'' (x) = 10 (1 + \ln (\frac{x}{4})) + 5 \cdot 1$

Step 2.2.3.3

Multiply $5$ by $1$ .

$f'' (x) = 10 (1 + \ln (\frac{x}{4})) + 5$

Step 2.2.4

Simplify.

Tap for more steps...

Step 2.2.4.1

Apply the distributive property.

$f'' (x) = 10 \cdot 1 + 10 \ln (\frac{x}{4}) + 5$

Step 2.2.4.2

Combine terms.

Tap for more steps...

Step 2.2.4.2.1

Multiply $10$ by $1$ .

$f'' (x) = 10 + 10 \ln (\frac{x}{4}) + 5$

Step 2.2.4.2.2

Add $10$ and $5$ .

$f'' (x) = 10 \ln (\frac{x}{4}) + 15$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $10 \ln (\frac{x}{4}) + 15$ .

$10 \ln (\frac{x}{4}) + 15$

Step 3

Set the second derivative equal to $0$ then solve the equation $10 \ln (\frac{x}{4}) + 15 = 0$ .

Tap for more steps...

Step 3.1

Set the second derivative equal to $0$ .

$10 \ln (\frac{x}{4}) + 15 = 0$

Step 3.2

Subtract $15$ from both sides of the equation.

$10 \ln (\frac{x}{4}) = - 15$

Step 3.3

Divide each term in $10 \ln (\frac{x}{4}) = - 15$ by $10$ and simplify.

Tap for more steps...

Step 3.3.1

Divide each term in $10 \ln (\frac{x}{4}) = - 15$ by $10$ .

$\frac{10 \ln (\frac{x}{4})}{10} = \frac{- 15}{10}$

Step 3.3.2

Simplify the left side.

Tap for more steps...

Step 3.3.2.1

Cancel the common factor of $10$ .

Tap for more steps...

Step 3.3.2.1.1

Cancel the common factor.

$\frac{10 \ln (\frac{x}{4})}{10} = \frac{- 15}{10}$

Step 3.3.2.1.2

Divide $\ln (\frac{x}{4})$ by $1$ .

$\ln (\frac{x}{4}) = \frac{- 15}{10}$

Step 3.3.3

Simplify the right side.

Tap for more steps...

Step 3.3.3.1

Cancel the common factor of $- 15$ and $10$ .

Tap for more steps...

Step 3.3.3.1.1

Factor $5$ out of $- 15$ .

$\ln (\frac{x}{4}) = \frac{5 (- 3)}{10}$

Step 3.3.3.1.2

Cancel the common factors.

Tap for more steps...

Step 3.3.3.1.2.1

Factor $5$ out of $10$ .

$\ln (\frac{x}{4}) = \frac{5 \cdot - 3}{5 \cdot 2}$

Step 3.3.3.1.2.2

Cancel the common factor.

$\ln (\frac{x}{4}) = \frac{5 \cdot - 3}{5 \cdot 2}$

Step 3.3.3.1.2.3

Rewrite the expression.

$\ln (\frac{x}{4}) = \frac{- 3}{2}$

Step 3.3.3.2

Move the negative in front of the fraction.

$\ln (\frac{x}{4}) = - \frac{3}{2}$

Step 3.4

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (\frac{x}{4})} = e^{- \frac{3}{2}}$

Step 3.5

Rewrite $\ln (\frac{x}{4}) = - \frac{3}{2}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- \frac{3}{2}} = \frac{x}{4}$

Step 3.6

Solve for $x$ .

Tap for more steps...

Step 3.6.1

Rewrite the equation as $\frac{x}{4} = e^{- \frac{3}{2}}$ .

$\frac{x}{4} = e^{- \frac{3}{2}}$

Step 3.6.2

Multiply both sides of the equation by $4$ .

$4 \frac{x}{4} = 4 e^{- \frac{3}{2}}$

Step 3.6.3

Simplify both sides of the equation.

Tap for more steps...

Step 3.6.3.1

Simplify the left side.

Tap for more steps...

Step 3.6.3.1.1

Cancel the common factor of $4$ .

Tap for more steps...

Step 3.6.3.1.1.1

Cancel the common factor.

$4 \frac{x}{4} = 4 e^{- \frac{3}{2}}$

Step 3.6.3.1.1.2

Rewrite the expression.

$x = 4 e^{- \frac{3}{2}}$

Step 3.6.3.2

Simplify the right side.

Tap for more steps...

Step 3.6.3.2.1

Simplify $4 e^{- \frac{3}{2}}$ .

Tap for more steps...

Step 3.6.3.2.1.1

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

$x = 4 \frac{1}{e^{\frac{3}{2}}}$

Step 3.6.3.2.1.2

Combine $4$ and $\frac{1}{e^{\frac{3}{2}}}$ .

$x = \frac{4}{e^{\frac{3}{2}}}$

Step 4

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

Tap for more steps...

Step 4.1

Substitute $\frac{4}{e^{\frac{3}{2}}}$ in $f (x) = 5 x^{2} \ln (\frac{x}{4})$ to find the value of $y$ .

Tap for more steps...

Step 4.1.1

Replace the variable $x$ with $\frac{4}{e^{\frac{3}{2}}}$ in the expression.

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 {(\frac{4}{e^{\frac{3}{2}}})}^{2} \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2

Simplify the result.

Tap for more steps...

Step 4.1.2.1

Apply the product rule to $\frac{4}{e^{\frac{3}{2}}}$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 (\frac{4^{2}}{{(e^{\frac{3}{2}})}^{2}}) \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.2

Raise $4$ to the power of $2$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 (\frac{16}{{(e^{\frac{3}{2}})}^{2}}) \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.3

Multiply the exponents in ${(e^{\frac{3}{2}})}^{2}$ .

Tap for more steps...

Step 4.1.2.3.1

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

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 (\frac{16}{e^{\frac{3}{2} \cdot 2}}) \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.3.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.2.3.2.1

Cancel the common factor.

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 (\frac{16}{e^{\frac{3}{2} \cdot 2}}) \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.3.2.2

Rewrite the expression.

$f (\frac{4}{e^{\frac{3}{2}}}) = 5 (\frac{16}{e^{3}}) \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.4

Multiply $5 \frac{16}{e^{3}}$ .

Tap for more steps...

Step 4.1.2.4.1

Combine $5$ and $\frac{16}{e^{3}}$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{5 \cdot 16}{e^{3}} \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.4.2

Multiply $5$ by $16$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (\frac{\frac{4}{e^{\frac{3}{2}}}}{4})$

Step 4.1.2.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (\frac{4}{e^{\frac{3}{2}}} \cdot \frac{1}{4})$

Step 4.1.2.6

Combine.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (\frac{4 \cdot 1}{e^{\frac{3}{2}} \cdot 4})$

Step 4.1.2.7

Reduce the expression $\frac{4 \cdot 1}{e^{\frac{3}{2}} \cdot 4}$ by cancelling the common factors.

Tap for more steps...

Step 4.1.2.7.1

Cancel the common factor.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (\frac{4 \cdot 1}{e^{\frac{3}{2}} \cdot 4})$

Step 4.1.2.7.2

Rewrite the expression.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (\frac{1}{e^{\frac{3}{2}}})$

Step 4.1.2.8

Move $e^{\frac{3}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \ln (e^{- \frac{3}{2}})$

Step 4.1.2.9

Expand $\ln (e^{- \frac{3}{2}})$ by moving $- \frac{3}{2}$ outside the logarithm.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot (- \frac{3}{2} \cdot \ln (e))$

Step 4.1.2.10

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

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot (- \frac{3}{2} \cdot 1)$

Step 4.1.2.11

Multiply $- 1$ by $1$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot (- \frac{3}{2})$

Step 4.1.2.12

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.2.12.1

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{80}{e^{3}} \cdot \frac{- 3}{2}$

Step 4.1.2.12.2

Factor $2$ out of $80$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{2 (40)}{e^{3}} \cdot \frac{- 3}{2}$

Step 4.1.2.12.3

Cancel the common factor.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{2 \cdot 40}{e^{3}} \cdot \frac{- 3}{2}$

Step 4.1.2.12.4

Rewrite the expression.

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{40}{e^{3}} \cdot - 3$

Step 4.1.2.13

Combine $\frac{40}{e^{3}}$ and $- 3$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{40 \cdot - 3}{e^{3}}$

Step 4.1.2.14

Simplify the expression.

Tap for more steps...

Step 4.1.2.14.1

Multiply $40$ by $- 3$ .

$f (\frac{4}{e^{\frac{3}{2}}}) = \frac{- 120}{e^{3}}$

Step 4.1.2.14.2

Move the negative in front of the fraction.

$f (\frac{4}{e^{\frac{3}{2}}}) = - \frac{120}{e^{3}}$

Step 4.1.2.15

The final answer is $- \frac{120}{e^{3}}$ .

$- \frac{120}{e^{3}}$

Step 4.2

The point found by substituting $\frac{4}{e^{\frac{3}{2}}}$ in $f (x) = 5 x^{2} \ln (\frac{x}{4})$ is $(\frac{4}{e^{\frac{3}{2}}}, - \frac{120}{e^{3}})$ . This point can be an inflection point.

$(\frac{4}{e^{\frac{3}{2}}}, - \frac{120}{e^{3}})$

Step 5

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

$(- \infty, \frac{4}{e^{\frac{3}{2}}}) \cup (\frac{4}{e^{\frac{3}{2}}}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{4}{e^{\frac{3}{2}}})$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $0.79252064$ in the expression.

$f'' (0.79252064) = 10 \ln (\frac{0.79252064}{4}) + 15$

Step 6.2

Simplify the result.

Tap for more steps...

Step 6.2.1

Simplify each term.

Tap for more steps...

Step 6.2.1.1

Divide $0.79252064$ by $4$ .

$f'' (0.79252064) = 10 \ln (0.19813016) + 15$

Step 6.2.1.2

Simplify $10 \ln (0.19813016)$ by moving $10$ inside the logarithm.

$f'' (0.79252064) = \ln ({0.19813016}^{10}) + 15$

Step 6.2.1.3

Raise $0.19813016$ to the power of $10$ .

$f'' (0.79252064) = \ln (0.00000009) + 15$

Step 6.2.2

The final answer is $\ln (0.00000009) + 15$ .

$\ln (0.00000009) + 15$

Step 6.3

At $0.79252064$ , the second derivative is $- 1.18831089$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{4}{e^{\frac{3}{2}}})$

Decreasing on $(- \infty, \frac{4}{e^{\frac{3}{2}}})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{4}{e^{\frac{3}{2}}}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Tap for more steps...

Step 7.1

Replace the variable $x$ with $0.99252064$ in the expression.

$f'' (0.99252064) = 10 \ln (\frac{0.99252064}{4}) + 15$

Step 7.2

Simplify the result.

Tap for more steps...

Step 7.2.1

Simplify each term.

Tap for more steps...

Step 7.2.1.1

Divide $0.99252064$ by $4$ .

$f'' (0.99252064) = 10 \ln (0.24813016) + 15$

Step 7.2.1.2

Simplify $10 \ln (0.24813016)$ by moving $10$ inside the logarithm.

$f'' (0.99252064) = \ln ({0.24813016}^{10}) + 15$

Step 7.2.1.3

Raise $0.24813016$ to the power of $10$ .

$f'' (0.99252064) = \ln (0.00000088) + 15$

Step 7.2.2

The final answer is $\ln (0.00000088) + 15$ .

$\ln (0.00000088) + 15$

Step 7.3

At $0.99252064$ , the second derivative is $1.06198168$ . Since this is positive, the second derivative is increasing on the interval $(\frac{4}{e^{\frac{3}{2}}}, \infty)$ .

Increasing on $(\frac{4}{e^{\frac{3}{2}}}, \infty)$ 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 $(\frac{4}{e^{\frac{3}{2}}}, - \frac{120}{e^{3}})$ .

$(\frac{4}{e^{\frac{3}{2}}}, - \frac{120}{e^{3}})$

Step 9