Calculus Examples

$x e^{x}$

Step 1

Write $x e^{x}$ as a function.

$f (x) = x e^{x}$

Step 2

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

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

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

$x \frac{d}{d x} [e^{x}] + e^{x} \frac{d}{d x} [x]$

Step 2.1.1.2

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

$x e^{x} + e^{x} \frac{d}{d x} [x]$

Step 2.1.1.3

Differentiate using the Power Rule.

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Step 2.1.1.3.1

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

$x e^{x} + e^{x} \cdot 1$

Step 2.1.1.3.2

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

$f' (x) = x e^{x} + e^{x}$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

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

$\frac{d}{d x} [x e^{x}] + \frac{d}{d x} [e^{x}]$

Step 2.1.2.2

Evaluate $\frac{d}{d x} [x e^{x}]$ .

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Step 2.1.2.2.1

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

$x \frac{d}{d x} [e^{x}] + e^{x} \frac{d}{d x} [x] + \frac{d}{d x} [e^{x}]$

Step 2.1.2.2.2

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

$x e^{x} + e^{x} \frac{d}{d x} [x] + \frac{d}{d x} [e^{x}]$

Step 2.1.2.2.3

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

$x e^{x} + e^{x} \cdot 1 + \frac{d}{d x} [e^{x}]$

Step 2.1.2.2.4

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

$x e^{x} + e^{x} + \frac{d}{d x} [e^{x}]$

Step 2.1.2.3

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

$x e^{x} + e^{x} + e^{x}$

Step 2.1.2.4

Simplify.

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Step 2.1.2.4.1

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

$x e^{x} + 2 e^{x}$

Step 2.1.2.4.2

Reorder terms.

$e^{x} x + 2 e^{x}$

Step 2.1.2.4.3

Reorder factors in $e^{x} x + 2 e^{x}$ .

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

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $x e^{x} + 2 e^{x}$ .

$x e^{x} + 2 e^{x}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $x e^{x} + 2 e^{x} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$x e^{x} + 2 e^{x} = 0$

Step 2.2.2

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

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Step 2.2.2.1

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

$e^{x} x + 2 e^{x} = 0$

Step 2.2.2.2

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

$e^{x} x + e^{x} \cdot 2 = 0$

Step 2.2.2.3

Factor $e^{x}$ out of $e^{x} x + e^{x} \cdot 2$ .

$e^{x} (x + 2) = 0$

Step 2.2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{x} = 0$

$x + 2 = 0$

Step 2.2.4

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

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Step 2.2.4.1

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

$e^{x} = 0$

Step 2.2.4.2

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

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Step 2.2.4.2.1

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

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

Step 2.2.4.2.2

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

Undefined

Step 2.2.4.2.3

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

No solution

Step 2.2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 2.2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.2.6

The final solution is all the values that make $e^{x} (x + 2) = 0$ true.

$x = - 2$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

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

$(- \infty, - 2) \cup (- 2, \infty)$

Step 5

Substitute any number from the interval $(- \infty, - 2)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

Replace the variable $x$ with $- 4$ in the expression.

$f'' (- 4) = (- 4) \cdot e^{- 4} + 2 e^{- 4}$

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

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

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

Step 5.2.1.2

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

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

Step 5.2.1.3

Move the negative in front of the fraction.

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

Step 5.2.1.4

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

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

Step 5.2.1.5

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

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

Step 5.2.2

Combine fractions.

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Step 5.2.2.1

Combine the numerators over the common denominator.

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

Step 5.2.2.2

Simplify the expression.

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Step 5.2.2.2.1

Add $- 4$ and $2$ .

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

Step 5.2.2.2.2

Move the negative in front of the fraction.

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

Step 5.2.3

The final answer is $- \frac{2}{e^{4}}$ .

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

Step 5.3

The graph is concave down on the interval $(- \infty, - 2)$ because $f'' (- 4)$ is negative.

Concave down on $(- \infty, - 2)$ since $f'' (x)$ is negative

Step 6

Substitute any number from the interval $(- 2, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = (0) \cdot e^{0} + 2 e^{0}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Anything raised to $0$ is $1$ .

$f'' (0) = 0 \cdot 1 + 2 e^{0}$

Step 6.2.1.2

Multiply $0$ by $1$ .

$f'' (0) = 0 + 2 e^{0}$

Step 6.2.1.3

Anything raised to $0$ is $1$ .

$f'' (0) = 0 + 2 \cdot 1$

Step 6.2.1.4

Multiply $2$ by $1$ .

$f'' (0) = 0 + 2$

Step 6.2.2

Add $0$ and $2$ .

$f'' (0) = 2$

Step 6.2.3

The final answer is $2$ .

$2$

Step 6.3

The graph is concave up on the interval $(- 2, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- 2, \infty)$ since $f'' (x)$ is positive

Step 7

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

Concave down on $(- \infty, - 2)$ since $f'' (x)$ is negative

Concave up on $(- 2, \infty)$ since $f'' (x)$ is positive

Step 8