Calculus Examples

$f (x) = x^{4} e^{x} - 5$

Step 1

Find the first derivative of the function.

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By the Sum Rule, the derivative of $x^{4} e^{x} - 5$ with respect to $x$ is $\frac{d}{d x} [x^{4} e^{x}] + \frac{d}{d x} [- 5]$ .

$\frac{d}{d x} [x^{4} e^{x}] + \frac{d}{d x} [- 5]$

Evaluate $\frac{d}{d x} [x^{4} e^{x}]$ .

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

$x^{4} \frac{d}{d x} [e^{x}] + e^{x} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 5]$

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

$x^{4} e^{x} + e^{x} \frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 5]$

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

$x^{4} e^{x} + e^{x} (4 x^{3}) + \frac{d}{d x} [- 5]$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$x^{4} e^{x} + e^{x} (4 x^{3}) + 0$

Simplify.

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Add $x^{4} e^{x} + e^{x} (4 x^{3})$ and $0$ .

$x^{4} e^{x} + e^{x} (4 x^{3})$

Reorder terms.

$e^{x} x^{4} + 4 e^{x} x^{3}$

Reorder factors in $e^{x} x^{4} + 4 e^{x} x^{3}$ .

$x^{4} e^{x} + 4 x^{3} e^{x}$

Step 2

Find the second derivative of the function.

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

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

Evaluate $\frac{d}{d x} [x^{4} e^{x}]$ .

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

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

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

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

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

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

Evaluate $\frac{d}{d x} [4 x^{3} e^{x}]$ .

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

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

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

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

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

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

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

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Multiply $3$ by $4$ .

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

Add $e^{x} (4 x^{3})$ and $4 x^{3} e^{x}$ .

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Move $e^{x}$ .

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

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

$f'' (x) = x^{4} e^{x} + 8 x^{3} e^{x} + 12 e^{x} x^{2}$

Reorder terms.

$f'' (x) = e^{x} x^{4} + 8 e^{x} x^{3} + 12 e^{x} x^{2}$

Reorder factors in $e^{x} x^{4} + 8 e^{x} x^{3} + 12 e^{x} x^{2}$ .

$f'' (x) = x^{4} e^{x} + 8 x^{3} e^{x} + 12 x^{2} e^{x}$

Step 3

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

$x^{4} e^{x} + 4 x^{3} e^{x} = 0$

Step 4

Find the first derivative.

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Find the first derivative.

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By the Sum Rule, the derivative of $x^{4} e^{x} - 5$ with respect to $x$ is $\frac{d}{d x} [x^{4} e^{x}] + \frac{d}{d x} [- 5]$ .

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

Evaluate $\frac{d}{d x} [x^{4} e^{x}]$ .

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

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

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

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

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

$f' (x) = x^{4} e^{x} + e^{x} (4 x^{3}) + \frac{d}{d x} (- 5)$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$f' (x) = x^{4} e^{x} + e^{x} (4 x^{3}) + 0$

Simplify.

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Add $x^{4} e^{x} + e^{x} (4 x^{3})$ and $0$ .

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

Reorder terms.

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

Reorder factors in $e^{x} x^{4} + 4 e^{x} x^{3}$ .

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

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

$x^{4} e^{x} + 4 x^{3} e^{x}$

Step 5

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

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Set the first derivative equal to $0$ .

$x^{4} e^{x} + 4 x^{3} e^{x} = 0$

Factor $x^{3} e^{x}$ out of $x^{4} e^{x} + 4 x^{3} e^{x}$ .

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Factor $x^{3} e^{x}$ out of $x^{4} e^{x}$ .

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

Factor $x^{3} e^{x}$ out of $4 x^{3} e^{x}$ .

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

Factor $x^{3} e^{x}$ out of $x^{3} e^{x} (x) + x^{3} e^{x} (4)$ .

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

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

$x^{3} = 0$

$e^{x} = 0$

$x + 4 = 0$

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Solve $x^{3} = 0$ for $x$ .

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Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

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Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

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

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

$e^{x} = 0$

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

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

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

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

Undefined

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

No solution

Set $x + 4$ equal to $0$ and solve for $x$ .

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Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

The final solution is all the values that make $x^{3} e^{x} (x + 4) = 0$ true.

$x = 0, - 4$

Step 6

Find the values where the derivative is undefined.

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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.

Step 7

Critical points to evaluate.

$x = 0, - 4$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(0)}^{4} e^{0} + 8 {(0)}^{3} e^{0} + 12 {(0)}^{2} e^{0}$

Step 9

Evaluate the second derivative.

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Simplify each term.

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Raising $0$ to any positive power yields $0$ .

$0 e^{0} + 8 {(0)}^{3} e^{0} + 12 {(0)}^{2} e^{0}$

Anything raised to $0$ is $1$ .

$0 \cdot 1 + 8 {(0)}^{3} e^{0} + 12 {(0)}^{2} e^{0}$

Multiply $0$ by $1$ .

$0 + 8 {(0)}^{3} e^{0} + 12 {(0)}^{2} e^{0}$

Raising $0$ to any positive power yields $0$ .

$0 + 8 \cdot 0 e^{0} + 12 {(0)}^{2} e^{0}$

Multiply $8$ by $0$ .

$0 + 0 e^{0} + 12 {(0)}^{2} e^{0}$

Anything raised to $0$ is $1$ .

$0 + 0 \cdot 1 + 12 {(0)}^{2} e^{0}$

Multiply $0$ by $1$ .

$0 + 0 + 12 {(0)}^{2} e^{0}$

Raising $0$ to any positive power yields $0$ .

$0 + 0 + 12 \cdot 0 e^{0}$

Multiply $12$ by $0$ .

$0 + 0 + 0 e^{0}$

Anything raised to $0$ is $1$ .

$0 + 0 + 0 \cdot 1$

Multiply $0$ by $1$ .

$0 + 0 + 0$

Simplify by adding numbers.

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Add $0$ and $0$ .

$0 + 0$

Add $0$ and $0$ .

$0$

Step 10

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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

$(- \infty, - 4) \cup (- 4, 0) \cup (0, \infty)$

Substitute any number, such as $- 6$ , from the interval $(- \infty, - 4)$ in the first derivative $x^{4} e^{x} + 4 x^{3} e^{x}$ to check if the result is negative or positive.

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Replace the variable $x$ with $- 6$ in the expression.

$f' (- 6) = {(- 6)}^{4} e^{- 6} + 4 {(- 6)}^{3} e^{- 6}$

Simplify the result.

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Simplify each term.

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Raise $- 6$ to the power of $4$ .

$f' (- 6) = 1296 e^{- 6} + 4 {(- 6)}^{3} e^{- 6}$

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

$f' (- 6) = 1296 (\frac{1}{e^{6}}) + 4 {(- 6)}^{3} e^{- 6}$

Combine $1296$ and $\frac{1}{e^{6}}$ .

$f' (- 6) = \frac{1296}{e^{6}} + 4 {(- 6)}^{3} e^{- 6}$

Raise $- 6$ to the power of $3$ .

$f' (- 6) = \frac{1296}{e^{6}} + 4 \cdot (- 216 e^{- 6})$

Multiply $4$ by $- 216$ .

$f' (- 6) = \frac{1296}{e^{6}} - 864 e^{- 6}$

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

$f' (- 6) = \frac{1296}{e^{6}} - 864 \frac{1}{e^{6}}$

Combine $- 864$ and $\frac{1}{e^{6}}$ .

$f' (- 6) = \frac{1296}{e^{6}} + \frac{- 864}{e^{6}}$

Move the negative in front of the fraction.

$f' (- 6) = \frac{1296}{e^{6}} - \frac{864}{e^{6}}$

Combine fractions.

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Combine the numerators over the common denominator.

$f' (- 6) = \frac{1296 - 864}{e^{6}}$

Subtract $864$ from $1296$ .

$f' (- 6) = \frac{432}{e^{6}}$

The final answer is $\frac{432}{e^{6}}$ .

$\frac{432}{e^{6}}$

Substitute any number, such as $- 2$ , from the interval $(- 4, 0)$ in the first derivative $x^{4} e^{x} + 4 x^{3} e^{x}$ to check if the result is negative or positive.

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Replace the variable $x$ with $- 2$ in the expression.

$f' (- 2) = {(- 2)}^{4} e^{- 2} + 4 {(- 2)}^{3} e^{- 2}$

Simplify the result.

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Simplify each term.

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

$f' (- 2) = 16 e^{- 2} + 4 {(- 2)}^{3} e^{- 2}$

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

$f' (- 2) = 16 (\frac{1}{e^{2}}) + 4 {(- 2)}^{3} e^{- 2}$

Combine $16$ and $\frac{1}{e^{2}}$ .

$f' (- 2) = \frac{16}{e^{2}} + 4 {(- 2)}^{3} e^{- 2}$

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

$f' (- 2) = \frac{16}{e^{2}} + 4 \cdot (- 8 e^{- 2})$

Multiply $4$ by $- 8$ .

$f' (- 2) = \frac{16}{e^{2}} - 32 e^{- 2}$

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

$f' (- 2) = \frac{16}{e^{2}} - 32 \frac{1}{e^{2}}$

Combine $- 32$ and $\frac{1}{e^{2}}$ .

$f' (- 2) = \frac{16}{e^{2}} + \frac{- 32}{e^{2}}$

Move the negative in front of the fraction.

$f' (- 2) = \frac{16}{e^{2}} - \frac{32}{e^{2}}$

Combine fractions.

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Combine the numerators over the common denominator.

$f' (- 2) = \frac{16 - 32}{e^{2}}$

Simplify the expression.

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Subtract $32$ from $16$ .

$f' (- 2) = \frac{- 16}{e^{2}}$

Move the negative in front of the fraction.

$f' (- 2) = - \frac{16}{e^{2}}$

The final answer is $- \frac{16}{e^{2}}$ .

$- \frac{16}{e^{2}}$

Substitute any number, such as $2$ , from the interval $(0, \infty)$ in the first derivative $x^{4} e^{x} + 4 x^{3} e^{x}$ to check if the result is negative or positive.

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Replace the variable $x$ with $2$ in the expression.

$f' (2) = {(2)}^{4} e^{2} + 4 {(2)}^{3} e^{2}$

Simplify the result.

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Simplify each term.

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

$f' (2) = 16 e^{2} + 4 {(2)}^{3} e^{2}$

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

$f' (2) = 16 e^{2} + 4 \cdot (8 e^{2})$

Multiply $4$ by $8$ .

$f' (2) = 16 e^{2} + 32 e^{2}$

Add $16 e^{2}$ and $32 e^{2}$ .

$f' (2) = 48 e^{2}$

The final answer is $48 e^{2}$ .

$48 e^{2}$

Since the first derivative changed signs from positive to negative around $x = - 4$ , then $x = - 4$ is a local maximum.

$x = - 4$ is a local maximum

Since the first derivative changed signs from negative to positive around $x = 0$ , then $x = 0$ is a local minimum.

$x = 0$ is a local minimum

These are the local extrema for $f (x) = x^{4} e^{x} - 5$ .

$x = - 4$ is a local maximum

$x = 0$ is a local minimum

$x = - 4$ is a local maximum

$x = 0$ is a local minimum

Step 11