Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 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 $g (x) = e^{x}$ .

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder terms.

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

Step 1.1.4.2

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

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

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

$x = 0$

$e^{x} = 0$

$x + 2 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

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

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Step 2.5.1

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

$e^{x} = 0$

Step 2.5.2

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

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Step 2.5.2.1

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

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

Step 2.5.2.2

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

Undefined

Step 2.5.2.3

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

No solution

Step 2.6

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

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Step 2.6.1

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

$x + 2 = 0$

Step 2.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.7

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

$x = 0, - 2$

Step 3

The values which make the derivative equal to $0$ are $0, - 2$ .

$0, - 2$

Step 4

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the derivative $0$ or undefined.

$(- \infty, - 2) \cup (- 2, 0) \cup (0, \infty)$

Step 5

Substitute a value from the interval $(- \infty, - 2)$ into the derivative to determine if the function is increasing or decreasing.

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Step 5.1

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

$f' (- 3) = {(- 3)}^{2} e^{- 3} + 2 (- 3) \cdot e^{- 3}$

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

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

$f' (- 3) = 9 e^{- 3} + 2 (- 3) \cdot e^{- 3}$

Step 5.2.1.2

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

$f' (- 3) = 9 (\frac{1}{e^{3}}) + 2 (- 3) \cdot e^{- 3}$

Step 5.2.1.3

Combine $9$ and $\frac{1}{e^{3}}$ .

$f' (- 3) = \frac{9}{e^{3}} + 2 (- 3) \cdot e^{- 3}$

Step 5.2.1.4

Multiply $2$ by $- 3$ .

$f' (- 3) = \frac{9}{e^{3}} - 6 \cdot e^{- 3}$

Step 5.2.1.5

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

$f' (- 3) = \frac{9}{e^{3}} - 6 \cdot \frac{1}{e^{3}}$

Step 5.2.1.6

Combine $- 6$ and $\frac{1}{e^{3}}$ .

$f' (- 3) = \frac{9}{e^{3}} + \frac{- 6}{e^{3}}$

Step 5.2.1.7

Move the negative in front of the fraction.

$f' (- 3) = \frac{9}{e^{3}} - \frac{6}{e^{3}}$

Step 5.2.2

Combine fractions.

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Step 5.2.2.1

Combine the numerators over the common denominator.

$f' (- 3) = \frac{9 - 6}{e^{3}}$

Step 5.2.2.2

Subtract $6$ from $9$ .

$f' (- 3) = \frac{3}{e^{3}}$

Step 5.2.3

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

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

Step 5.3

At $x = - 3$ the derivative is $\frac{3}{e^{3}}$ . Since this is positive, the function is increasing on $(- \infty, - 2)$ .

Increasing on $(- \infty, - 2)$ since $f' (x) > 0$

Step 6

Substitute a value from the interval $(- 2, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

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

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

Raise $- 1$ to the power of $2$ .

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

Step 6.2.1.2

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

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

Step 6.2.1.3

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

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

Step 6.2.1.4

Multiply $2$ by $- 1$ .

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

Step 6.2.1.5

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

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

Step 6.2.1.6

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

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

Step 6.2.1.7

Move the negative in front of the fraction.

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

Step 6.2.2

Combine fractions.

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Step 6.2.2.1

Combine the numerators over the common denominator.

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

Step 6.2.2.2

Simplify the expression.

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Step 6.2.2.2.1

Subtract $2$ from $1$ .

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

Step 6.2.2.2.2

Move the negative in front of the fraction.

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

Step 6.2.3

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

$- \frac{1}{e}$

Step 6.3

At $x = - 1$ the derivative is $- \frac{1}{e}$ . Since this is negative, the function is decreasing on $(- 2, 0)$ .

Decreasing on $(- 2, 0)$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

One to any power is one.

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

Step 7.2.1.2

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

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

Step 7.2.1.3

Simplify.

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

Step 7.2.1.4

Multiply $2$ by $1$ .

$f' (1) = e + 2 e$

Step 7.2.2

Add $e$ and $2 e$ .

$f' (1) = 3 e$

Step 7.2.3

The final answer is $3 e$ .

$3 e$

Step 7.3

At $x = 1$ the derivative is $3 e$ . Since this is positive, the function is increasing on $(0, \infty)$ .

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

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, - 2), (0, \infty)$

Decreasing on: $(- 2, 0)$

Step 9