Calculus Examples

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

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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 $f (x) = x^{2}$ and $g (x) = \ln (x)$ .

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

Step 1.1.1.2

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

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

Step 1.1.1.3

Differentiate using the Power Rule.

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Step 1.1.1.3.1

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

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

Step 1.1.1.3.2

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

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Step 1.1.1.3.2.1

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

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

Step 1.1.1.3.2.2

Cancel the common factors.

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Step 1.1.1.3.2.2.1

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

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

Step 1.1.1.3.2.2.2

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

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

Step 1.1.1.3.2.2.3

Cancel the common factor.

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

Step 1.1.1.3.2.2.4

Rewrite the expression.

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

Step 1.1.1.3.2.2.5

Divide $x$ by $1$ .

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

Step 1.1.1.3.3

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

$f' (x) = x + \ln (x) (2 x)$

Step 1.1.1.3.4

Reorder terms.

$f' (x) = 2 x \ln (x) + x$

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

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

Step 1.1.2.2

Evaluate $\frac{d}{d x} [2 x \ln (x)]$ .

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

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

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

Step 1.1.2.2.3

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

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

Step 1.1.2.2.4

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

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

Step 1.1.2.2.5

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

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

Step 1.1.2.2.6

Cancel the common factor of $x$ .

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Step 1.1.2.2.6.1

Cancel the common factor.

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

Step 1.1.2.2.6.2

Rewrite the expression.

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

Step 1.1.2.2.7

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

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

Step 1.1.2.3

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

$f'' (x) = 2 (1 + \ln (x)) + 1$

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

Apply the distributive property.

$f'' (x) = 2 \cdot 1 + 2 \ln (x) + 1$

Step 1.1.2.4.2

Combine terms.

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Step 1.1.2.4.2.1

Multiply $2$ by $1$ .

$f'' (x) = 2 + 2 \ln (x) + 1$

Step 1.1.2.4.2.2

Add $2$ and $1$ .

$f'' (x) = 2 \ln (x) + 3$

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $2 \ln (x) + 3$ .

$2 \ln (x) + 3$

Step 1.2

Set the second derivative equal to $0$ then solve the equation $2 \ln (x) + 3 = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$2 \ln (x) + 3 = 0$

Step 1.2.2

Subtract $3$ from both sides of the equation.

$2 \ln (x) = - 3$

Step 1.2.3

Divide each term in $2 \ln (x) = - 3$ by $2$ and simplify.

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Step 1.2.3.1

Divide each term in $2 \ln (x) = - 3$ by $2$ .

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

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $2$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

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

Step 1.2.3.2.1.2

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

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

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Move the negative in front of the fraction.

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

Step 1.2.4

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

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

Step 1.2.5

Rewrite $\ln (x) = - \frac{3}{2}$ 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 $b^{y} = x$ .

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

Step 1.2.6

Solve for $x$ .

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Step 1.2.6.1

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

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

Step 1.2.6.2

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

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

Step 2

Find the domain of $f (x) = x^{2} \ln (x)$ .

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Step 2.1

Set the argument in $\ln (x)$ greater than $0$ to find where the expression is defined.

$x > 0$

Step 2.2

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Interval Notation:

$(0, \infty)$

Set-Builder Notation:

${x | x > 0}$

Step 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(0, \frac{1}{e^{\frac{3}{2}}}) \cup (\frac{1}{e^{\frac{3}{2}}}, \infty)$

Step 4

Substitute any number from the interval $(0, \frac{1}{e^{\frac{3}{2}}})$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $0.11$ in the expression.

$f'' (0.11) = 2 \ln (0.11) + 3$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Simplify $2 \ln (0.11)$ by moving $2$ inside the logarithm.

$f'' (0.11) = \ln ({0.11}^{2}) + 3$

Step 4.2.1.2

Raise $0.11$ to the power of $2$ .

$f'' (0.11) = \ln (0.0121) + 3$

Step 4.2.2

The final answer is $\ln (0.0121) + 3$ .

$\ln (0.0121) + 3$

Step 4.3

The graph is concave down on the interval $(0, \frac{1}{e^{\frac{3}{2}}})$ because $f'' (0.11)$ is negative.

Concave down on $(0, \frac{1}{e^{\frac{3}{2}}})$ since $f'' (x)$ is negative

Step 5

Substitute any number from the interval $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $3$ in the expression.

$f'' (3) = 2 \ln (3) + 3$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Simplify $2 \ln (3)$ by moving $2$ inside the logarithm.

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

Step 5.2.1.2

Raise $3$ to the power of $2$ .

$f'' (3) = \ln (9) + 3$

Step 5.2.2

The final answer is $\ln (9) + 3$ .

$\ln (9) + 3$

Step 5.3

The graph is concave up on the interval $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ because $f'' (3)$ is positive.

Concave up on $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ since $f'' (x)$ is positive

Step 6

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave down on $(0, \frac{1}{e^{\frac{3}{2}}})$ since $f'' (x)$ is negative

Concave up on $(\frac{1}{e^{\frac{3}{2}}}, \infty)$ since $f'' (x)$ is positive

Step 7