Calculus Examples

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

Step 1

Write $e^{- \frac{x^{2}}{2}}$ as a function.

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

Step 2

Find the second derivative.

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Find the first derivative.

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

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Simplify terms.

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Multiply $2$ by $- 1$ .

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

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

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

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

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

Cancel the common factor of $- 2$ and $2$ .

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

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

Cancel the common factors.

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Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- e^{- \frac{x^{2}}{2}} x$ by $1$ .

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

Reorder factors in $- e^{- \frac{x^{2}}{2}} x$ .

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

Find the second derivative.

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

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

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 $f (x) = x$ and $g (x) = e^{- \frac{x^{2}}{2}}$ .

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

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

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

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

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

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

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

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

Differentiate.

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

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

Combine fractions.

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

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

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

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

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

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

Combine fractions.

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Multiply $2$ by $- 1$ .

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

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

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

Combine $\frac{- 2 (x e^{- \frac{x^{2}}{2}})}{2}$ and $x$ .

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

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

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

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

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

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

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

Reduce the expression by cancelling the common factors.

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Add $1$ and $1$ .

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

Cancel the common factor of $- 2$ and $2$ .

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

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

Cancel the common factors.

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Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x^{2} e^{- \frac{x^{2}}{2}}$ by $1$ .

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

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

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

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

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Multiply $- 1$ by $- 1$ .

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

Multiply $x^{2}$ by $1$ .

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

Reorder terms.

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

Reorder factors in $e^{- \frac{x^{2}}{2}} x^{2} - e^{- \frac{x^{2}}{2}}$ .

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

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

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

Step 3

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

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Set the second derivative equal to $0$ .

$x^{2} e^{- \frac{x^{2}}{2}} - e^{- \frac{x^{2}}{2}} = 0$

Factor the left side of the equation.

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

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

$e^{- \frac{x^{2}}{2}} x^{2} - e^{- \frac{x^{2}}{2}} = 0$

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

$e^{- \frac{x^{2}}{2}} x^{2} + e^{- \frac{x^{2}}{2}} \cdot - 1 = 0$

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

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

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

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

Factor.

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

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

Remove unnecessary parentheses.

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

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

$e^{- \frac{x^{2}}{2}} = 0$

$x + 1 = 0$

$x - 1 = 0$

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

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

$e^{- \frac{x^{2}}{2}} = 0$

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

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

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

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

Undefined

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

No solution

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

$x = - 1, 1$

Step 4

Find the points where the second derivative is $0$ .

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Substitute $- 1$ in $f (x) = e^{- \frac{x^{2}}{2}}$ to find the value of $y$ .

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

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

Simplify the result.

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

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

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

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

The final answer is $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{1}{e^{\frac{1}{2}}}$

The point found by substituting $- 1$ in $f (x) = e^{- \frac{x^{2}}{2}}$ is $(- 1, \frac{1}{e^{\frac{1}{2}}})$ . This point can be an inflection point.

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

Substitute $1$ in $f (x) = e^{- \frac{x^{2}}{2}}$ to find the value of $y$ .

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

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

Simplify the result.

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One to any power is one.

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

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

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

The final answer is $\frac{1}{e^{\frac{1}{2}}}$ .

$\frac{1}{e^{\frac{1}{2}}}$

The point found by substituting $1$ in $f (x) = e^{- \frac{x^{2}}{2}}$ is $(1, \frac{1}{e^{\frac{1}{2}}})$ . This point can be an inflection point.

$(1, \frac{1}{e^{\frac{1}{2}}})$

Determine the points that could be inflection points.

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

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, - 1) \cup (- 1, 1) \cup (1, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - 1)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $- 1.1$ in the expression.

$f'' (- 1.1) = {(- 1.1)}^{2} e^{- \frac{{(- 1.1)}^{2}}{2}} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Simplify the result.

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Simplify each term.

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Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = 1.21 e^{- \frac{{(- 1.1)}^{2}}{2}} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = 1.21 e^{- \frac{1.21}{2}} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Divide $1.21$ by $2$ .

$f'' (- 1.1) = 1.21 e^{- 1 \cdot 0.605} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Multiply $- 1$ by $0.605$ .

$f'' (- 1.1) = 1.21 e^{- 0.605} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

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

$f'' (- 1.1) = 1.21 (\frac{1}{e^{0.605}}) - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Combine $1.21$ and $\frac{1}{e^{0.605}}$ .

$f'' (- 1.1) = \frac{1.21}{e^{0.605}} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Replace $e$ with an approximation.

$f'' (- 1.1) = \frac{1.21}{{2.71828182}^{0.605}} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Raise $2.71828182$ to the power of $0.605$ .

$f'' (- 1.1) = \frac{1.21}{1.8312522} - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Divide $1.21$ by $1.8312522$ .

$f'' (- 1.1) = 0.66075005 - e^{- \frac{{(- 1.1)}^{2}}{2}}$

Raise $- 1.1$ to the power of $2$ .

$f'' (- 1.1) = 0.66075005 - e^{- \frac{1.21}{2}}$

Divide $1.21$ by $2$ .

$f'' (- 1.1) = 0.66075005 - e^{- 1 \cdot 0.605}$

Multiply $- 1$ by $0.605$ .

$f'' (- 1.1) = 0.66075005 - e^{- 0.605}$

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

$f'' (- 1.1) = 0.66075005 - \frac{1}{e^{0.605}}$

Subtract $\frac{1}{e^{0.605}}$ from $0.66075005$ .

$f'' (- 1.1) = 0.11467562$

The final answer is $0.11467562$ .

$0.11467562$

At $- 1.1$ , the second derivative is $0.11467562$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, - 1)$ .

Increasing on $(- \infty, - 1)$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(- 1, 1)$ into the second derivative to determine if it is increasing or decreasing.

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

$f'' (0) = {(0)}^{2} e^{- \frac{{(0)}^{2}}{2}} - e^{- \frac{{(0)}^{2}}{2}}$

Simplify the result.

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Simplify each term.

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

$f'' (0) = 0 e^{- \frac{{(0)}^{2}}{2}} - e^{- \frac{{(0)}^{2}}{2}}$

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

$f'' (0) = 0 e^{- \frac{0}{2}} - e^{- \frac{{(0)}^{2}}{2}}$

Divide $0$ by $2$ .

$f'' (0) = 0 e^{- 0} - e^{- \frac{{(0)}^{2}}{2}}$

Multiply $- 1$ by $0$ .

$f'' (0) = 0 e^{0} - e^{- \frac{{(0)}^{2}}{2}}$

Anything raised to $0$ is $1$ .

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

Multiply $0$ by $1$ .

$f'' (0) = 0 - e^{- \frac{{(0)}^{2}}{2}}$

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

$f'' (0) = 0 - e^{- \frac{0}{2}}$

Divide $0$ by $2$ .

$f'' (0) = 0 - e^{- 0}$

Multiply $- 1$ by $0$ .

$f'' (0) = 0 - e^{0}$

Anything raised to $0$ is $1$ .

$f'' (0) = 0 - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f'' (0) = 0 - 1$

Subtract $1$ from $0$ .

$f'' (0) = - 1$

The final answer is $- 1$ .

$- 1$

At $0$ , the second derivative is $- 1$ . Since this is negative, the second derivative is decreasing on the interval $(- 1, 1)$

Decreasing on $(- 1, 1)$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(1, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Replace the variable $x$ with $1.1$ in the expression.

$f'' (1.1) = {(1.1)}^{2} e^{- \frac{{(1.1)}^{2}}{2}} - e^{- \frac{{(1.1)}^{2}}{2}}$

Simplify the result.

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Simplify each term.

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Raise $1.1$ to the power of $2$ .

$f'' (1.1) = 1.21 e^{- \frac{{(1.1)}^{2}}{2}} - e^{- \frac{{(1.1)}^{2}}{2}}$

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = 1.21 e^{- \frac{1.21}{2}} - e^{- \frac{{(1.1)}^{2}}{2}}$

Divide $1.21$ by $2$ .

$f'' (1.1) = 1.21 e^{- 1 \cdot 0.605} - e^{- \frac{{(1.1)}^{2}}{2}}$

Multiply $- 1$ by $0.605$ .

$f'' (1.1) = 1.21 e^{- 0.605} - e^{- \frac{{(1.1)}^{2}}{2}}$

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

$f'' (1.1) = 1.21 (\frac{1}{e^{0.605}}) - e^{- \frac{{(1.1)}^{2}}{2}}$

Combine $1.21$ and $\frac{1}{e^{0.605}}$ .

$f'' (1.1) = \frac{1.21}{e^{0.605}} - e^{- \frac{{(1.1)}^{2}}{2}}$

Replace $e$ with an approximation.

$f'' (1.1) = \frac{1.21}{{2.71828182}^{0.605}} - e^{- \frac{{(1.1)}^{2}}{2}}$

Raise $2.71828182$ to the power of $0.605$ .

$f'' (1.1) = \frac{1.21}{1.8312522} - e^{- \frac{{(1.1)}^{2}}{2}}$

Divide $1.21$ by $1.8312522$ .

$f'' (1.1) = 0.66075005 - e^{- \frac{{(1.1)}^{2}}{2}}$

Raise $1.1$ to the power of $2$ .

$f'' (1.1) = 0.66075005 - e^{- \frac{1.21}{2}}$

Divide $1.21$ by $2$ .

$f'' (1.1) = 0.66075005 - e^{- 1 \cdot 0.605}$

Multiply $- 1$ by $0.605$ .

$f'' (1.1) = 0.66075005 - e^{- 0.605}$

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

$f'' (1.1) = 0.66075005 - \frac{1}{e^{0.605}}$

Subtract $\frac{1}{e^{0.605}}$ from $0.66075005$ .

$f'' (1.1) = 0.11467562$

The final answer is $0.11467562$ .

$0.11467562$

At $1.1$ , the second derivative is $0.11467562$ . Since this is positive, the second derivative is increasing on the interval $(1, \infty)$ .

Increasing on $(1, \infty)$ since $f'' (x) > 0$

Step 9

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection points in this case are $(- 1, \frac{1}{e^{\frac{1}{2}}}), (1, \frac{1}{e^{\frac{1}{2}}})$ .

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

Step 10