Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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 $f (x) = x^{4}$ and $g (x) = \ln (x)$ .

$x^{4} \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.2

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

$x^{4} \frac{1}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3

Differentiate using the Power Rule.

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Step 1.3.1

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

$\frac{x^{4}}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2

Cancel the common factor of $x^{4}$ and $x$ .

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Step 1.3.2.1

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

$\frac{x \cdot x^{3}}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2.2

Cancel the common factors.

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Step 1.3.2.2.1

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

$\frac{x \cdot x^{3}}{x^{1}} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2.2.2

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

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2.2.3

Cancel the common factor.

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2.2.4

Rewrite the expression.

$\frac{x^{3}}{1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.2.2.5

Divide $x^{3}$ by $1$ .

$x^{3} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$x^{3} + \ln (x) (4 x^{3})$

Step 1.3.4

Reorder terms.

$x^{3} + 4 x^{3} \ln (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x^{3} + 4 x^{3} \ln (x)$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x^{3} \ln (x)]$ .

$f'' (x) = \frac{d}{d x} (x^{3}) + \frac{d}{d x} (4 x^{3} \ln (x))$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 3 x^{2} + \frac{d}{d x} (4 x^{3} \ln (x))$

Step 2.2

Evaluate $\frac{d}{d x} [4 x^{3} \ln (x)]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3} \ln (x)$ with respect to $x$ is $4 \frac{d}{d x} [x^{3} \ln (x)]$ .

$f'' (x) = 3 x^{2} + 4 \frac{d}{d x} (x^{3} \ln (x))$

Step 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 $f (x) = x^{3}$ and $g (x) = \ln (x)$ .

$f'' (x) = 3 x^{2} + 4 (x^{3} \frac{d}{d x} (\ln (x)) + \ln (x) \frac{d}{d x} (x^{3}))$

Step 2.2.3

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

$f'' (x) = 3 x^{2} + 4 (x^{3} (\frac{1}{x}) + \ln (x) \frac{d}{d x} (x^{3}))$

Step 2.2.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 3 x^{2} + 4 (x^{3} (\frac{1}{x}) + \ln (x) (3 x^{2}))$

Step 2.2.5

Combine $x^{3}$ and $\frac{1}{x}$ .

$f'' (x) = 3 x^{2} + 4 (\frac{x^{3}}{x} + \ln (x) (3 x^{2}))$

Step 2.2.6

Cancel the common factor of $x^{3}$ and $x$ .

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Step 2.2.6.1

Factor $x$ out of $x^{3}$ .

$f'' (x) = 3 x^{2} + 4 (\frac{x \cdot x^{2}}{x} + \ln (x) (3 x^{2}))$

Step 2.2.6.2

Cancel the common factors.

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Step 2.2.6.2.1

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

$f'' (x) = 3 x^{2} + 4 (\frac{x \cdot x^{2}}{x} + \ln (x) (3 x^{2}))$

Step 2.2.6.2.2

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

$f'' (x) = 3 x^{2} + 4 (\frac{x \cdot x^{2}}{x \cdot 1} + \ln (x) (3 x^{2}))$

Step 2.2.6.2.3

Cancel the common factor.

$f'' (x) = 3 x^{2} + 4 (\frac{x \cdot x^{2}}{x \cdot 1} + \ln (x) (3 x^{2}))$

Step 2.2.6.2.4

Rewrite the expression.

$f'' (x) = 3 x^{2} + 4 (\frac{x^{2}}{1} + \ln (x) (3 x^{2}))$

Step 2.2.6.2.5

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

$f'' (x) = 3 x^{2} + 4 (x^{2} + \ln (x) (3 x^{2}))$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$f'' (x) = 3 x^{2} + 4 x^{2} + 4 (\ln (x) (3 x^{2}))$

Step 2.3.2

Combine terms.

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Step 2.3.2.1

Multiply $3$ by $4$ .

$f'' (x) = 3 x^{2} + 4 x^{2} + 12 (\ln (x) (x^{2}))$

Step 2.3.2.2

Add $3 x^{2}$ and $4 x^{2}$ .

$f'' (x) = 7 x^{2} + 12 \ln (x) x^{2}$

Step 2.3.3

Reorder terms.

$f'' (x) = 7 x^{2} + 12 x^{2} \ln (x)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$x^{3} + 4 x^{3} \ln (x) = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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 $f (x) = x^{4}$ and $g (x) = \ln (x)$ .

$x^{4} \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.2

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

$x^{4} \frac{1}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3

Differentiate using the Power Rule.

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Step 4.1.3.1

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

$\frac{x^{4}}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2

Cancel the common factor of $x^{4}$ and $x$ .

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Step 4.1.3.2.1

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

$\frac{x \cdot x^{3}}{x} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2.2

Cancel the common factors.

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Step 4.1.3.2.2.1

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

$\frac{x \cdot x^{3}}{x^{1}} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2.2.2

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

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2.2.3

Cancel the common factor.

$\frac{x \cdot x^{3}}{x \cdot 1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2.2.4

Rewrite the expression.

$\frac{x^{3}}{1} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.2.2.5

Divide $x^{3}$ by $1$ .

$x^{3} + \ln (x) \frac{d}{d x} [x^{4}]$

Step 4.1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$x^{3} + \ln (x) (4 x^{3})$

Step 4.1.3.4

Reorder terms.

$f' (x) = x^{3} + 4 x^{3} \ln (x)$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $x^{3} + 4 x^{3} \ln (x)$ .

$x^{3} + 4 x^{3} \ln (x)$

Step 5

Set the first derivative equal to $0$ then solve the equation $x^{3} + 4 x^{3} \ln (x) = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$x^{3} + 4 x^{3} \ln (x) = 0$

Step 5.2

Subtract $x^{3}$ from both sides of the equation.

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

Step 5.3

Divide each term in $4 x^{3} \ln (x) = - x^{3}$ by $4 x^{3}$ and simplify.

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Step 5.3.1

Divide each term in $4 x^{3} \ln (x) = - x^{3}$ by $4 x^{3}$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $x^{3}$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

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

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Cancel the common factor of $x^{3}$ .

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Step 5.3.3.1.1

Cancel the common factor.

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

Step 5.3.3.1.2

Rewrite the expression.

$\ln (x) = \frac{- 1}{4}$

Step 5.3.3.2

Move the negative in front of the fraction.

$\ln (x) = - \frac{1}{4}$

Step 5.4

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

$e^{\ln (x)} = e^{- \frac{1}{4}}$

Step 5.5

Rewrite $\ln (x) = - \frac{1}{4}$ 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 $b^{y} = x$ .

$e^{- \frac{1}{4}} = x$

Step 5.6

Solve for $x$ .

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Step 5.6.1

Rewrite the equation as $x = e^{- \frac{1}{4}}$ .

$x = e^{- \frac{1}{4}}$

Step 5.6.2

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

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

Step 6

Find the values where the derivative is undefined.

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Step 6.1

Set the argument in $\ln (x)$ less than or equal to $0$ to find where the expression is undefined.

$x \leq 0$

Step 6.2

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x \leq 0$

$(- \infty, 0]$

$x \leq 0$

$(- \infty, 0]$

Step 7

Critical points to evaluate.

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

Step 8

Evaluate the second derivative at $x = \frac{1}{e^{\frac{1}{4}}}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$7 {(\frac{1}{e^{\frac{1}{4}}})}^{2} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Apply the product rule to $\frac{1}{e^{\frac{1}{4}}}$ .

$7 \frac{1^{2}}{{(e^{\frac{1}{4}})}^{2}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.2

One to any power is one.

$7 \frac{1}{{(e^{\frac{1}{4}})}^{2}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.3

Multiply the exponents in ${(e^{\frac{1}{4}})}^{2}$ .

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Step 9.1.3.1

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

$7 \frac{1}{e^{\frac{1}{4} \cdot 2}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.3.2

Cancel the common factor of $2$ .

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Step 9.1.3.2.1

Factor $2$ out of $4$ .

$7 \frac{1}{e^{\frac{1}{2 (2)} \cdot 2}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.3.2.2

Cancel the common factor.

$7 \frac{1}{e^{\frac{1}{2 \cdot 2} \cdot 2}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.3.2.3

Rewrite the expression.

$7 \frac{1}{e^{\frac{1}{2}}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.4

Combine $7$ and $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{7}{e^{\frac{1}{2}}} + 12 {(\frac{1}{e^{\frac{1}{4}}})}^{2} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.5

Apply the product rule to $\frac{1}{e^{\frac{1}{4}}}$ .

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1^{2}}{{(e^{\frac{1}{4}})}^{2}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.6

One to any power is one.

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1}{{(e^{\frac{1}{4}})}^{2}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.7

Multiply the exponents in ${(e^{\frac{1}{4}})}^{2}$ .

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Step 9.1.7.1

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

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1}{e^{\frac{1}{4} \cdot 2}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.7.2

Cancel the common factor of $2$ .

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Step 9.1.7.2.1

Factor $2$ out of $4$ .

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1}{e^{\frac{1}{2 (2)} \cdot 2}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.7.2.2

Cancel the common factor.

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1}{e^{\frac{1}{2 \cdot 2} \cdot 2}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.7.2.3

Rewrite the expression.

$\frac{7}{e^{\frac{1}{2}}} + 12 \frac{1}{e^{\frac{1}{2}}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.8

Combine $12$ and $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 9.1.9

Move $e^{\frac{1}{4}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} \ln (e^{- \frac{1}{4}})$

Step 9.1.10

Expand $\ln (e^{- \frac{1}{4}})$ by moving $- \frac{1}{4}$ outside the logarithm.

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} (- \frac{1}{4} \ln (e))$

Step 9.1.11

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

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} (- \frac{1}{4} \cdot 1)$

Step 9.1.12

Multiply $- 1$ by $1$ .

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} (- \frac{1}{4})$

Step 9.1.13

Cancel the common factor of $4$ .

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Step 9.1.13.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$\frac{7}{e^{\frac{1}{2}}} + \frac{12}{e^{\frac{1}{2}}} \cdot \frac{- 1}{4}$

Step 9.1.13.2

Factor $4$ out of $12$ .

$\frac{7}{e^{\frac{1}{2}}} + \frac{4 (3)}{e^{\frac{1}{2}}} \cdot \frac{- 1}{4}$

Step 9.1.13.3

Cancel the common factor.

$\frac{7}{e^{\frac{1}{2}}} + \frac{4 \cdot 3}{e^{\frac{1}{2}}} \cdot \frac{- 1}{4}$

Step 9.1.13.4

Rewrite the expression.

$\frac{7}{e^{\frac{1}{2}}} + \frac{3}{e^{\frac{1}{2}}} \cdot - 1$

Step 9.1.14

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

$\frac{7}{e^{\frac{1}{2}}} + \frac{3 \cdot - 1}{e^{\frac{1}{2}}}$

Step 9.1.15

Multiply $3$ by $- 1$ .

$\frac{7}{e^{\frac{1}{2}}} + \frac{- 3}{e^{\frac{1}{2}}}$

Step 9.1.16

Move the negative in front of the fraction.

$\frac{7}{e^{\frac{1}{2}}} - \frac{3}{e^{\frac{1}{2}}}$

Step 9.2

Combine fractions.

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Step 9.2.1

Combine the numerators over the common denominator.

$\frac{7 - 3}{e^{\frac{1}{2}}}$

Step 9.2.2

Subtract $3$ from $7$ .

$\frac{4}{e^{\frac{1}{2}}}$

Step 10

$x = \frac{1}{e^{\frac{1}{4}}}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{1}{e^{\frac{1}{4}}}$ is a local minimum

Step 11

Find the y-value when $x = \frac{1}{e^{\frac{1}{4}}}$ .

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Step 11.1

Replace the variable $x$ with $\frac{1}{e^{\frac{1}{4}}}$ in the expression.

$f (\frac{1}{e^{\frac{1}{4}}}) = {(\frac{1}{e^{\frac{1}{4}}})}^{4} \ln (\frac{1}{e^{\frac{1}{4}}})$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify the expression.

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Step 11.2.1.1

Apply the product rule to $\frac{1}{e^{\frac{1}{4}}}$ .

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

Step 11.2.1.2

One to any power is one.

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

Step 11.2.2

Simplify the denominator.

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Step 11.2.2.1

Multiply the exponents in ${(e^{\frac{1}{4}})}^{4}$ .

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Step 11.2.2.1.1

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

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

Step 11.2.2.1.2

Cancel the common factor of $4$ .

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Step 11.2.2.1.2.1

Cancel the common factor.

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

Step 11.2.2.1.2.2

Rewrite the expression.

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

Step 11.2.2.2

Simplify.

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

Step 11.2.3

Move $e^{\frac{1}{4}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$f (\frac{1}{e^{\frac{1}{4}}}) = \frac{1}{e} \cdot \ln (e^{- \frac{1}{4}})$

Step 11.2.4

Expand $\ln (e^{- \frac{1}{4}})$ by moving $- \frac{1}{4}$ outside the logarithm.

$f (\frac{1}{e^{\frac{1}{4}}}) = \frac{1}{e} \cdot (- \frac{1}{4} \cdot \ln (e))$

Step 11.2.5

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

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

Step 11.2.6

Multiply $- 1$ by $1$ .

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

Step 11.2.7

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

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

Step 11.2.8

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

$f (\frac{1}{e^{\frac{1}{4}}}) = - \frac{1}{4 e}$

Step 11.2.9

The final answer is $- \frac{1}{4 e}$ .

$y = - \frac{1}{4 e}$

Step 12

These are the local extrema for $f (x) = x^{4} \ln (x)$ .

$(\frac{1}{e^{\frac{1}{4}}}, - \frac{1}{4 e})$ is a local minima

Step 13