Calculus Examples

$y = x^{4} e^{- x}$

Step 1

Write $y = x^{4} e^{- x}$ as a function.

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

Step 2

Find the first derivative of the function.

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Step 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 $f (x) = x^{4}$ and $g (x) = e^{- x}$ .

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

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- x$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $- x$ .

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

Step 2.3

Differentiate.

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 2.3.2

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

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

Step 2.3.3

Simplify the expression.

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Step 2.3.3.1

Multiply $- 1$ by $1$ .

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

Step 2.3.3.2

Move $- 1$ to the left of $e^{- x}$ .

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

Step 2.3.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 2.3.4

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

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

Step 2.4

Simplify.

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Step 2.4.1

Reorder terms.

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

Step 2.4.2

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

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

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{4} e^{- x}$ with respect to $x$ is $- \frac{d}{d x} [x^{4} e^{- x}]$ .

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

Step 3.2.2

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 $f (x) = x^{4}$ and $g (x) = e^{- x}$ .

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

Step 3.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 3.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$f'' (x) = - (x^{4} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - (x^{4} (e^{u_{1}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.3.3

Replace all occurrences of $u_{1}$ with $- x$ .

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

Step 3.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 3.2.5

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

$f'' (x) = - (x^{4} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x^{4})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.6

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

$f'' (x) = - (x^{4} (e^{- x} (- 1 \cdot 1)) + e^{- x} (4 x^{3})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.7

Multiply $- 1$ by $1$ .

$f'' (x) = - (x^{4} (e^{- x} \cdot - 1) + e^{- x} (4 x^{3})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.8

Move $- 1$ to the left of $e^{- x}$ .

$f'' (x) = - (x^{4} (- 1 \cdot e^{- x}) + e^{- x} (4 x^{3})) + \frac{d}{d x} (4 x^{3} e^{- x})$

Step 3.2.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u_{2}$ as $- x$ .

$f'' (x) = - (x^{4} (- e^{- x}) + e^{- x} (4 x^{3})) + 4 (x^{3} (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{3}))$

Step 3.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 3.3.3.3

Replace all occurrences of $u_{2}$ with $- x$ .

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

Step 3.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 3.3.5

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

$f'' (x) = - (x^{4} (- e^{- x}) + e^{- x} (4 x^{3})) + 4 (x^{3} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x^{3}))$

Step 3.3.6

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

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

Step 3.3.7

Multiply $- 1$ by $1$ .

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

Step 3.3.8

Move $- 1$ to the left of $e^{- x}$ .

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

Step 3.3.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

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

Step 3.4.2

Apply the distributive property.

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

Step 3.4.3

Combine terms.

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Step 3.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.3.2

Multiply $x^{4}$ by $1$ .

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

Step 3.4.3.3

Multiply $4$ by $- 1$ .

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

Step 3.4.3.4

Multiply $- 1$ by $4$ .

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

Step 3.4.3.5

Multiply $3$ by $4$ .

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

Step 3.4.3.6

Subtract $4 x^{3} e^{- x}$ from $- 4 e^{- x} x^{3}$ .

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Step 3.4.3.6.1

Move $e^{- x}$ .

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

Step 3.4.3.6.2

Subtract $4 x^{3} e^{- x}$ from $- 4 x^{3} e^{- x}$ .

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

Step 3.4.4

Reorder terms.

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

Step 3.4.5

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 4

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 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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

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

Step 5.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

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Step 5.1.2.1

To apply the Chain Rule, set $u$ as $- x$ .

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Replace all occurrences of $u$ with $- x$ .

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

Step 5.1.3

Differentiate.

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Step 5.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 5.1.3.2

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

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

Step 5.1.3.3

Simplify the expression.

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Step 5.1.3.3.1

Multiply $- 1$ by $1$ .

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

Step 5.1.3.3.2

Move $- 1$ to the left of $e^{- x}$ .

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

Step 5.1.3.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 5.1.3.4

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

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

Step 5.1.4

Simplify.

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Step 5.1.4.1

Reorder terms.

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

Step 5.1.4.2

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

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

Step 5.2

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 6

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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Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

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

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

Step 6.2.3

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

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

Step 6.3

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

$x^{3} = 0$

$e^{- x} = 0$

$- x + 4 = 0$

Step 6.4

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

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Step 6.4.1

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

$x^{3} = 0$

Step 6.4.2

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

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Step 6.4.2.1

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

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

Step 6.4.2.2

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

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Step 6.4.2.2.1

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

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

Step 6.4.2.2.2

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

$x = 0$

Step 6.5

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

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Step 6.5.1

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

$e^{- x} = 0$

Step 6.5.2

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

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Step 6.5.2.1

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

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

Step 6.5.2.2

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

Undefined

Step 6.5.2.3

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

No solution

Step 6.6

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

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Step 6.6.1

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

$- x + 4 = 0$

Step 6.6.2

Solve $- x + 4 = 0$ for $x$ .

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Step 6.6.2.1

Subtract $4$ from both sides of the equation.

$- x = - 4$

Step 6.6.2.2

Divide each term in $- x = - 4$ by $- 1$ and simplify.

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Step 6.6.2.2.1

Divide each term in $- x = - 4$ by $- 1$ .

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

Step 6.6.2.2.2

Simplify the left side.

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Step 6.6.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 6.6.2.2.2.2

Divide $x$ by $1$ .

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

Step 6.6.2.2.3

Simplify the right side.

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Step 6.6.2.2.3.1

Divide $- 4$ by $- 1$ .

$x = 4$

Step 6.7

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

$x = 0, 4$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 8

Critical points to evaluate.

$x = 0, 4$

Step 9

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 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

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

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

Step 10.1.2

Multiply $- 1$ by $0$ .

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

Step 10.1.3

Anything raised to $0$ is $1$ .

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

Step 10.1.4

Multiply $0$ by $1$ .

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

Step 10.1.5

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

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

Step 10.1.6

Multiply $- 8$ by $0$ .

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

Step 10.1.7

Multiply $- 1$ by $0$ .

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

Step 10.1.8

Anything raised to $0$ is $1$ .

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

Step 10.1.9

Multiply $0$ by $1$ .

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

Step 10.1.10

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

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

Step 10.1.11

Multiply $12$ by $0$ .

$0 + 0 + 0 e^{- (0)}$

Step 10.1.12

Multiply $- 1$ by $0$ .

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

Step 10.1.13

Anything raised to $0$ is $1$ .

$0 + 0 + 0 \cdot 1$

Step 10.1.14

Multiply $0$ by $1$ .

$0 + 0 + 0$

Step 10.2

Simplify by adding numbers.

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Step 10.2.1

Add $0$ and $0$ .

$0 + 0$

Step 10.2.2

Add $0$ and $0$ .

$0$

Step 11

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

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Step 11.1

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

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

Step 11.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ 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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Step 11.2.1

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

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

Step 11.2.2

Simplify the result.

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Step 11.2.2.1

Simplify each term.

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Step 11.2.2.1.1

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

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

Step 11.2.2.1.2

Multiply $- 1$ by $16$ .

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

Step 11.2.2.1.3

Multiply $- 1$ by $- 2$ .

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

Step 11.2.2.1.4

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

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

Step 11.2.2.1.5

Multiply $4$ by $- 8$ .

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

Step 11.2.2.1.6

Multiply $- 1$ by $- 2$ .

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

Step 11.2.2.2

Subtract $32 e^{2}$ from $- 16 e^{2}$ .

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

Step 11.2.2.3

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

$- 48 e^{2}$

Step 11.3

Substitute any number, such as $2$ , from the interval $(0, 4)$ 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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Step 11.3.1

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

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

Step 11.3.2

Simplify the result.

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Step 11.3.2.1

Simplify each term.

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Step 11.3.2.1.1

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

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

Step 11.3.2.1.2

Multiply $- 1$ by $16$ .

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

Step 11.3.2.1.3

Multiply $- 1$ by $2$ .

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

Step 11.3.2.1.4

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

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

Step 11.3.2.1.5

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

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

Step 11.3.2.1.6

Move the negative in front of the fraction.

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

Step 11.3.2.1.7

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

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

Step 11.3.2.1.8

Multiply $4$ by $8$ .

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

Step 11.3.2.1.9

Multiply $- 1$ by $2$ .

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

Step 11.3.2.1.10

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

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

Step 11.3.2.1.11

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

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

Step 11.3.2.2

Combine fractions.

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Step 11.3.2.2.1

Combine the numerators over the common denominator.

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

Step 11.3.2.2.2

Add $- 16$ and $32$ .

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

Step 11.3.2.3

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

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

Step 11.4

Substitute any number, such as $6$ , from the interval $(4, \infty)$ 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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Step 11.4.1

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

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

Step 11.4.2

Simplify the result.

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Step 11.4.2.1

Simplify each term.

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Step 11.4.2.1.1

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

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

Step 11.4.2.1.2

Multiply $- 1$ by $1296$ .

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

Step 11.4.2.1.3

Multiply $- 1$ by $6$ .

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

Step 11.4.2.1.4

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

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

Step 11.4.2.1.5

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

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

Step 11.4.2.1.6

Move the negative in front of the fraction.

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

Step 11.4.2.1.7

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

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

Step 11.4.2.1.8

Multiply $4$ by $216$ .

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

Step 11.4.2.1.9

Multiply $- 1$ by $6$ .

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

Step 11.4.2.1.10

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

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

Step 11.4.2.1.11

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

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

Step 11.4.2.2

Combine fractions.

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Step 11.4.2.2.1

Combine the numerators over the common denominator.

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

Step 11.4.2.2.2

Simplify the expression.

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Step 11.4.2.2.2.1

Add $- 1296$ and $864$ .

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

Step 11.4.2.2.2.2

Move the negative in front of the fraction.

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

Step 11.4.2.3

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

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

Step 11.5

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

Step 11.6

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

Step 11.7

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

$x = 0$ is a local minimum

$x = 4$ is a local maximum

$x = 0$ is a local minimum

$x = 4$ is a local maximum

Step 12