Calculus Examples

$f (x) = x e^{- 0.5 x^{2}}$

Step 1

Find the first derivative of the function.

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Step 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 $g (x) = e^{- 0.5 x^{2}}$ .

$x \frac{d}{d x} [e^{- 0.5 x^{2}}] + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 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 $g (x) = - 0.5 x^{2}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $- 0.5 x^{2}$ .

$x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.2.2

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

$x (e^{u} \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.2.3

Replace all occurrences of $u$ with $- 0.5 x^{2}$ .

$x (e^{- 0.5 x^{2}} \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.3

Differentiate.

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Step 1.3.1

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x^{2}$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x^{2}]$ .

$x (e^{- 0.5 x^{2}} (- 0.5 \frac{d}{d x} [x^{2}])) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.3.2

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

$x (e^{- 0.5 x^{2}} (- 0.5 (2 x))) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.3.3

Multiply $2$ by $- 0.5$ .

$x (e^{- 0.5 x^{2}} (- x)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.4

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

$x^{1} x (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.5

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

$x^{1} x^{1} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{1 + 1} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.7

Simplify the expression.

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Step 1.7.1

Add $1$ and $1$ .

$x^{2} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.7.2

Move $- 1$ to the left of $e^{- 0.5 x^{2}}$ .

$x^{2} (- 1 \cdot e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.7.3

Rewrite $- 1 e^{- 0.5 x^{2}}$ as $- e^{- 0.5 x^{2}}$ .

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 1.8

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

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \cdot 1$

Step 1.9

Multiply $e^{- 0.5 x^{2}}$ by $1$ .

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}}$

Step 1.10

Simplify.

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Step 1.10.1

Reorder terms.

$- e^{- 0.5 x^{2}} x^{2} + e^{- 0.5 x^{2}}$

Step 1.10.2

Reorder factors in $- e^{- 0.5 x^{2}} x^{2} + e^{- 0.5 x^{2}}$ .

$- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$ with respect to $x$ is $\frac{d}{d x} [- x^{2} e^{- 0.5 x^{2}}] + \frac{d}{d x} [e^{- 0.5 x^{2}}]$ .

$f'' (x) = \frac{d}{d x} (- x^{2} e^{- 0.5 x^{2}}) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2

Evaluate $\frac{d}{d x} [- x^{2} e^{- 0.5 x^{2}}]$ .

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2} e^{- 0.5 x^{2}}$ with respect to $x$ is $- \frac{d}{d x} [x^{2} e^{- 0.5 x^{2}}]$ .

$f'' (x) = - \frac{d}{d x} (x^{2} e^{- 0.5 x^{2}}) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 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 $f (x) = x^{2}$ and $g (x) = e^{- 0.5 x^{2}}$ .

$f'' (x) = - (x^{2} \frac{d}{d x} (e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.3

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 $g (x) = - 0.5 x^{2}$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- 0.5 x^{2}$ .

$f'' (x) = - (x^{2} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- 0.5 x^{2})) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.3.2

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

$f'' (x) = - (x^{2} (e^{u_{1}} \frac{d}{d x} (- 0.5 x^{2})) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $- 0.5 x^{2}$ .

$f'' (x) = - (x^{2} (e^{- 0.5 x^{2}} \frac{d}{d x} (- 0.5 x^{2})) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.4

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x^{2}$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - (x^{2} (e^{- 0.5 x^{2}} (- 0.5 \frac{d}{d x} x^{2})) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.5

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

$f'' (x) = - (x^{2} (e^{- 0.5 x^{2}} (- 0.5 (2 x))) + e^{- 0.5 x^{2}} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.6

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

$f'' (x) = - (x^{2} (e^{- 0.5 x^{2}} (- 0.5 (2 x))) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.7

Multiply $2$ by $- 0.5$ .

$f'' (x) = - (x^{2} (e^{- 0.5 x^{2}} (- x)) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.8

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.2.8.1

Move $x$ .

$f'' (x) = - (x \cdot x^{2} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.8.2

Multiply $x$ by $x^{2}$ .

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Step 2.2.8.2.1

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

$f'' (x) = - (x \cdot x^{2} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.8.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = - (x^{1 + 2} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.8.3

Add $1$ and $2$ .

$f'' (x) = - (x^{3} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.9

Move $- 1$ to the left of $e^{- 0.5 x^{2}}$ .

$f'' (x) = - (x^{3} (- 1 \cdot e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.2.10

Rewrite $- 1 e^{- 0.5 x^{2}}$ as $- e^{- 0.5 x^{2}}$ .

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d x} (e^{- 0.5 x^{2}})$

Step 2.3

Evaluate $\frac{d}{d x} [e^{- 0.5 x^{2}}]$ .

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Step 2.3.1

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 $g (x) = - 0.5 x^{2}$ .

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Step 2.3.1.1

To apply the Chain Rule, set $u_{2}$ as $- 0.5 x^{2}$ .

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + \frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- 0.5 x^{2})$

Step 2.3.1.2

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

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + e^{u_{2}} \frac{d}{d x} (- 0.5 x^{2})$

Step 2.3.1.3

Replace all occurrences of $u_{2}$ with $- 0.5 x^{2}$ .

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} \frac{d}{d x} (- 0.5 x^{2})$

Step 2.3.2

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x^{2}$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- 0.5 \frac{d}{d x} x^{2})$

Step 2.3.3

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

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- 0.5 (2 x))$

Step 2.3.4

Multiply $2$ by $- 0.5$ .

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- x)$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = - (x^{3} (- e^{- 0.5 x^{2}})) - (e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- x)$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (x^{3} (e^{- 0.5 x^{2}})) - (e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- x)$

Step 2.4.2.2

Multiply $x^{3}$ by $1$ .

$f'' (x) = x^{3} e^{- 0.5 x^{2}} - (e^{- 0.5 x^{2}} (2 x)) + e^{- 0.5 x^{2}} (- x)$

Step 2.4.2.3

Multiply $2$ by $- 1$ .

$f'' (x) = x^{3} e^{- 0.5 x^{2}} - 2 (e^{- 0.5 x^{2}} (x)) + e^{- 0.5 x^{2}} (- x)$

Step 2.4.2.4

Move $e^{- 0.5 x^{2}}$ .

$f'' (x) = x^{3} e^{- 0.5 x^{2}} - 2 x e^{- 0.5 x^{2}} - x e^{- 0.5 x^{2}}$

Step 2.4.2.5

Subtract $x e^{- 0.5 x^{2}}$ from $- 2 x e^{- 0.5 x^{2}}$ .

$f'' (x) = x^{3} e^{- 0.5 x^{2}} - 3 x e^{- 0.5 x^{2}}$

Step 2.4.3

Reorder terms.

$f'' (x) = e^{- 0.5 x^{2}} x^{3} - 3 e^{- 0.5 x^{2}} x$

Step 2.4.4

Reorder factors in $e^{- 0.5 x^{2}} x^{3} - 3 e^{- 0.5 x^{2}} x$ .

$f'' (x) = x^{3} e^{- 0.5 x^{2}} - 3 x e^{- 0.5 x^{2}}$

Step 3

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

$- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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 $g (x) = e^{- 0.5 x^{2}}$ .

$x \frac{d}{d x} [e^{- 0.5 x^{2}}] + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.2

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 $g (x) = - 0.5 x^{2}$ .

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Step 4.1.2.1

To apply the Chain Rule, set $u$ as $- 0.5 x^{2}$ .

$x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.2.2

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

$x (e^{u} \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.2.3

Replace all occurrences of $u$ with $- 0.5 x^{2}$ .

$x (e^{- 0.5 x^{2}} \frac{d}{d x} [- 0.5 x^{2}]) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.3

Differentiate.

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Step 4.1.3.1

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x^{2}$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x^{2}]$ .

$x (e^{- 0.5 x^{2}} (- 0.5 \frac{d}{d x} [x^{2}])) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.3.2

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

$x (e^{- 0.5 x^{2}} (- 0.5 (2 x))) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.3.3

Multiply $2$ by $- 0.5$ .

$x (e^{- 0.5 x^{2}} (- x)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.4

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

$x^{1} x (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.5

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

$x^{1} x^{1} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{1 + 1} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.7

Simplify the expression.

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Step 4.1.7.1

Add $1$ and $1$ .

$x^{2} (e^{- 0.5 x^{2}} \cdot (- 1)) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.7.2

Move $- 1$ to the left of $e^{- 0.5 x^{2}}$ .

$x^{2} (- 1 \cdot e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.7.3

Rewrite $- 1 e^{- 0.5 x^{2}}$ as $- e^{- 0.5 x^{2}}$ .

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \frac{d}{d x} [x]$

Step 4.1.8

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

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}} \cdot 1$

Step 4.1.9

Multiply $e^{- 0.5 x^{2}}$ by $1$ .

$x^{2} (- e^{- 0.5 x^{2}}) + e^{- 0.5 x^{2}}$

Step 4.1.10

Simplify.

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Step 4.1.10.1

Reorder terms.

$- e^{- 0.5 x^{2}} x^{2} + e^{- 0.5 x^{2}}$

Step 4.1.10.2

Reorder factors in $- e^{- 0.5 x^{2}} x^{2} + e^{- 0.5 x^{2}}$ .

$f' (x) = - x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$ .

$- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}} = 0$

Step 5.2

Factor the left side of the equation.

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Step 5.2.1

Factor $e^{- 0.5 x^{2}}$ out of $- x^{2} e^{- 0.5 x^{2}} + e^{- 0.5 x^{2}}$ .

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Step 5.2.1.1

Factor $e^{- 0.5 x^{2}}$ out of $- x^{2} e^{- 0.5 x^{2}}$ .

$e^{- 0.5 x^{2}} (- x^{2}) + e^{- 0.5 x^{2}} = 0$

Step 5.2.1.2

Multiply by $1$ .

$e^{- 0.5 x^{2}} (- x^{2}) + e^{- 0.5 x^{2}} \cdot 1 = 0$

Step 5.2.1.3

Factor $e^{- 0.5 x^{2}}$ out of $e^{- 0.5 x^{2}} (- x^{2}) + e^{- 0.5 x^{2}} \cdot 1$ .

$e^{- 0.5 x^{2}} (- x^{2} + 1) = 0$

Step 5.2.2

Rewrite $1$ as $1^{2}$ .

$e^{- 0.5 x^{2}} (- x^{2} + 1^{2}) = 0$

Step 5.2.3

Reorder $- x^{2}$ and $1^{2}$ .

$e^{- 0.5 x^{2}} (1^{2} - x^{2}) = 0$

Step 5.2.4

Factor.

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Step 5.2.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$e^{- 0.5 x^{2}} ((1 + x) (1 - x)) = 0$

Step 5.2.4.2

Remove unnecessary parentheses.

$e^{- 0.5 x^{2}} (1 + x) (1 - x) = 0$

Step 5.3

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

$e^{- 0.5 x^{2}} = 0$

$1 + x = 0$

$1 - x = 0$

Step 5.4

Set $e^{- 0.5 x^{2}}$ equal to $0$ and solve for $x$ .

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Step 5.4.1

Set $e^{- 0.5 x^{2}}$ equal to $0$ .

$e^{- 0.5 x^{2}} = 0$

Step 5.4.2

Solve $e^{- 0.5 x^{2}} = 0$ for $x$ .

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Step 5.4.2.1

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

$\ln (e^{- 0.5 x^{2}}) = \ln (0)$

Step 5.4.2.2

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

Undefined

Step 5.4.2.3

There is no solution for $e^{- 0.5 x^{2}} = 0$

No solution

Step 5.5

Set $1 + x$ equal to $0$ and solve for $x$ .

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Step 5.5.1

Set $1 + x$ equal to $0$ .

$1 + x = 0$

Step 5.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.6

Set $1 - x$ equal to $0$ and solve for $x$ .

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Step 5.6.1

Set $1 - x$ equal to $0$ .

$1 - x = 0$

Step 5.6.2

Solve $1 - x = 0$ for $x$ .

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Step 5.6.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 5.6.2.2

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

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Step 5.6.2.2.1

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

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

Step 5.6.2.2.2

Simplify the left side.

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Step 5.6.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.6.2.2.2.2

Divide $x$ by $1$ .

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

Step 5.6.2.2.3

Simplify the right side.

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Step 5.6.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 5.7

The final solution is all the values that make $e^{- 0.5 x^{2}} (1 + x) (1 - x) = 0$ true.

$x = - 1, 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

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 7

Critical points to evaluate.

$x = - 1, 1$

Step 8

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(- 1)}^{3} e^{- 0.5 {(- 1)}^{2}} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Raise $- 1$ to the power of $3$ .

$- 1 e^{- 0.5 {(- 1)}^{2}} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.2

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

$- 1 e^{- 0.5 \cdot 1} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.3

Multiply $- 0.5$ by $1$ .

$- 1 e^{- 0.5} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.4

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

$- 1 \frac{1}{e^{0.5}} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.5

Rewrite $- 1 \frac{1}{e^{0.5}}$ as $- \frac{1}{e^{0.5}}$ .

$- \frac{1}{e^{0.5}} - 3 \cdot - 1 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.6

Multiply $- 3$ by $- 1$ .

$- \frac{1}{e^{0.5}} + 3 e^{- 0.5 {(- 1)}^{2}}$

Step 9.1.7

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

$- \frac{1}{e^{0.5}} + 3 e^{- 0.5 \cdot 1}$

Step 9.1.8

Multiply $- 0.5$ by $1$ .

$- \frac{1}{e^{0.5}} + 3 e^{- 0.5}$

Step 9.1.9

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

$- \frac{1}{e^{0.5}} + 3 \frac{1}{e^{0.5}}$

Step 9.1.10

Combine $3$ and $\frac{1}{e^{0.5}}$ .

$- \frac{1}{e^{0.5}} + \frac{3}{e^{0.5}}$

Step 9.2

Combine fractions.

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Step 9.2.1

Combine the numerators over the common denominator.

$\frac{- 1 + 3}{e^{0.5}}$

Step 9.2.2

Add $- 1$ and $3$ .

$\frac{2}{e^{0.5}}$

Step 10

$x = - 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = - 1$ is a local minimum

Step 11

Find the y-value when $x = - 1$ .

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Step 11.1

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

$f (- 1) = (- 1) \cdot e^{- 0.5 {(- 1)}^{2}}$

Step 11.2

Simplify the result.

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Step 11.2.1

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

$f (- 1) = - 1 \cdot e^{- 0.5 \cdot 1}$

Step 11.2.2

Multiply $- 0.5$ by $1$ .

$f (- 1) = - 1 \cdot e^{- 0.5}$

Step 11.2.3

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

$f (- 1) = - 1 \cdot \frac{1}{e^{0.5}}$

Step 11.2.4

Rewrite $- 1 \frac{1}{e^{0.5}}$ as $- \frac{1}{e^{0.5}}$ .

$f (- 1) = - \frac{1}{e^{0.5}}$

Step 11.2.5

The final answer is $- \frac{1}{e^{0.5}}$ .

$y = - \frac{1}{e^{0.5}}$

Step 12

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(1)}^{3} e^{- 0.5 {(1)}^{2}} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

One to any power is one.

$1 e^{- 0.5 {(1)}^{2}} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13.1.2

Multiply $e^{- 0.5 {(1)}^{2}}$ by $1$ .

$e^{- 0.5 {(1)}^{2}} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13.1.3

One to any power is one.

$e^{- 0.5 \cdot 1} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13.1.4

Multiply $- 0.5$ by $1$ .

$e^{- 0.5} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13.1.5

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

$\frac{1}{e^{0.5}} - 3 \cdot 1 e^{- 0.5 {(1)}^{2}}$

Step 13.1.6

Multiply $- 3$ by $1$ .

$\frac{1}{e^{0.5}} - 3 e^{- 0.5 {(1)}^{2}}$

Step 13.1.7

One to any power is one.

$\frac{1}{e^{0.5}} - 3 e^{- 0.5 \cdot 1}$

Step 13.1.8

Multiply $- 0.5$ by $1$ .

$\frac{1}{e^{0.5}} - 3 e^{- 0.5}$

Step 13.1.9

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

$\frac{1}{e^{0.5}} - 3 \frac{1}{e^{0.5}}$

Step 13.1.10

Combine $- 3$ and $\frac{1}{e^{0.5}}$ .

$\frac{1}{e^{0.5}} + \frac{- 3}{e^{0.5}}$

Step 13.1.11

Move the negative in front of the fraction.

$\frac{1}{e^{0.5}} - \frac{3}{e^{0.5}}$

Step 13.2

Combine fractions.

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Step 13.2.1

Combine the numerators over the common denominator.

$\frac{1 - 3}{e^{0.5}}$

Step 13.2.2

Simplify the expression.

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Step 13.2.2.1

Subtract $3$ from $1$ .

$\frac{- 2}{e^{0.5}}$

Step 13.2.2.2

Move the negative in front of the fraction.

$- \frac{2}{e^{0.5}}$

Step 14

$x = 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 1$ is a local maximum

Step 15

Find the y-value when $x = 1$ .

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Step 15.1

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

$f (1) = (1) \cdot e^{- 0.5 {(1)}^{2}}$

Step 15.2

Simplify the result.

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Step 15.2.1

Multiply $e^{- 0.5 {(1)}^{2}}$ by $1$ .

$f (1) = e^{- 0.5 {(1)}^{2}}$

Step 15.2.2

One to any power is one.

$f (1) = e^{- 0.5 \cdot 1}$

Step 15.2.3

Multiply $- 0.5$ by $1$ .

$f (1) = e^{- 0.5}$

Step 15.2.4

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

$f (1) = \frac{1}{e^{0.5}}$

Step 15.2.5

The final answer is $\frac{1}{e^{0.5}}$ .

$y = \frac{1}{e^{0.5}}$

Step 16

These are the local extrema for $f (x) = x e^{- 0.5 x^{2}}$ .

$(- 1, - \frac{1}{e^{0.5}})$ is a local minima

$(1, \frac{1}{e^{0.5}})$ is a local maxima

Step 17