Calculus Examples

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

Step 1

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

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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

Step 2.1.2

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

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Step 2.1.2.1

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 $g (x) = - x$ .

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

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

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

Step 2.1.2.1.3

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.2.4

Multiply $- 1$ by $1$ .

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

Step 2.1.2.5

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

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

Step 2.1.2.6

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

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

Step 2.1.3

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

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Step 2.1.3.1

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

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 2.1.3.3.1

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

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

Step 2.1.3.3.2

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

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

Step 2.1.3.3.3

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

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

Step 2.1.3.4

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

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

Step 2.1.3.5

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

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

Step 2.1.3.6

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

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

Step 2.1.3.7

Multiply $- 1$ by $1$ .

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

Step 2.1.3.8

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

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

Step 2.1.3.9

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

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

Step 2.1.3.10

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

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

Step 2.1.4

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

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Step 2.1.4.1

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

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

Step 2.1.4.2

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 2.1.4.2.1

To apply the Chain Rule, set $u_{3}$ as $- x$ .

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

Step 2.1.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

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

Step 2.1.4.2.3

Replace all occurrences of $u_{3}$ with $- x$ .

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

Step 2.1.4.3

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

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

Step 2.1.4.4

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

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

Step 2.1.4.5

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

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

Step 2.1.4.6

Multiply $- 1$ by $1$ .

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

Step 2.1.4.7

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

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

Step 2.1.4.8

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

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

Step 2.1.5

Simplify.

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Step 2.1.5.1

Apply the distributive property.

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

Step 2.1.5.2

Combine terms.

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Step 2.1.5.2.1

Multiply $- 1$ by $2$ .

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

Step 2.1.5.2.2

Add $- e^{- x}$ and $2 e^{- x}$ .

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

Step 2.1.5.2.3

Add $- 2 x e^{- x}$ and $e^{- x} (2 x)$ .

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Step 2.1.5.2.3.1

Move $e^{- x}$ .

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

Step 2.1.5.2.3.2

Add $- 2 x e^{- x}$ and $2 x e^{- x}$ .

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

Step 2.1.5.2.4

Add $e^{- x} + x^{2} (- e^{- x})$ and $0$ .

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

Step 2.1.5.3

Reorder terms.

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

Step 2.1.5.4

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

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

Step 2.2

Find the second derivative.

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Step 2.2.1

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

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

Step 2.2.2

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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 $g (x) = - x$ .

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Step 2.2.2.3.1

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

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

Step 2.2.2.3.2

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

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

Step 2.2.2.3.3

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

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

Step 2.2.2.4

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

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

Step 2.2.2.5

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

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

Step 2.2.2.6

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

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

Step 2.2.2.7

Multiply $- 1$ by $1$ .

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

Step 2.2.2.8

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

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

Step 2.2.2.9

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

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

Step 2.2.3

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

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Step 2.2.3.1

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 $g (x) = - x$ .

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Step 2.2.3.1.1

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

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

Step 2.2.3.1.2

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

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

Step 2.2.3.1.3

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.3.4

Multiply $- 1$ by $1$ .

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

Step 2.2.3.5

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

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

Step 2.2.3.6

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

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

Step 2.2.4

Simplify.

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Step 2.2.4.1

Apply the distributive property.

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

Step 2.2.4.2

Combine terms.

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Step 2.2.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 2.2.4.2.2

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

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

Step 2.2.4.2.3

Multiply $2$ by $- 1$ .

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

Step 2.2.4.3

Reorder terms.

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

Step 2.2.4.4

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

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

Step 2.3

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

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

Step 3

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

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.3

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

$e^{- x} = 0$

$x^{2} - 2 x - 1 = 0$

Step 3.4

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

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Step 3.4.1

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

$e^{- x} = 0$

Step 3.4.2

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

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Step 3.4.2.1

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

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

Step 3.4.2.2

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

Undefined

Step 3.4.2.3

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

No solution

Step 3.5

Set $x^{2} - 2 x - 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x^{2} - 2 x - 1$ equal to $0$ .

$x^{2} - 2 x - 1 = 0$

Step 3.5.2

Solve $x^{2} - 2 x - 1 = 0$ for $x$ .

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Step 3.5.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3.5.2.2

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 3.5.2.3

Simplify.

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Step 3.5.2.3.1

Simplify the numerator.

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Step 3.5.2.3.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.3.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.3.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 3.5.2.3.1.3

Add $4$ and $4$ .

$x = \frac{2 \pm \sqrt{8}}{2 \cdot 1}$

Step 3.5.2.3.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.5.2.3.1.4.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 3.5.2.3.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 3.5.2.3.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 3.5.2.3.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{2}}{2}$

Step 3.5.2.3.3

Simplify $\frac{2 \pm 2 \sqrt{2}}{2}$ .

$x = 1 \pm \sqrt{2}$

Step 3.5.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 3.5.2.4.1

Simplify the numerator.

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Step 3.5.2.4.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.4.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.5.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.4.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 3.5.2.4.1.3

Add $4$ and $4$ .

$x = \frac{2 \pm \sqrt{8}}{2 \cdot 1}$

Step 3.5.2.4.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.5.2.4.1.4.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 3.5.2.4.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 3.5.2.4.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 3.5.2.4.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{2}}{2}$

Step 3.5.2.4.3

Simplify $\frac{2 \pm 2 \sqrt{2}}{2}$ .

$x = 1 \pm \sqrt{2}$

Step 3.5.2.4.4

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{2}$

Step 3.5.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 3.5.2.5.1

Simplify the numerator.

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Step 3.5.2.5.1.1

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

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Step 3.5.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1}}{2 \cdot 1}$

Step 3.5.2.5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4}}{2 \cdot 1}$

Step 3.5.2.5.1.3

Add $4$ and $4$ .

$x = \frac{2 \pm \sqrt{8}}{2 \cdot 1}$

Step 3.5.2.5.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 3.5.2.5.1.4.1

Factor $4$ out of $8$ .

$x = \frac{2 \pm \sqrt{4 (2)}}{2 \cdot 1}$

Step 3.5.2.5.1.4.2

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Step 3.5.2.5.1.5

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{2}}{2 \cdot 1}$

Step 3.5.2.5.2

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{2}}{2}$

Step 3.5.2.5.3

Simplify $\frac{2 \pm 2 \sqrt{2}}{2}$ .

$x = 1 \pm \sqrt{2}$

Step 3.5.2.5.4

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{2}$

Step 3.5.2.6

The final answer is the combination of both solutions.

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 3.6

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

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

Step 4

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

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Step 4.1

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

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Step 4.1.1

Replace the variable $x$ with $1 + \sqrt{2}$ in the expression.

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

Apply the distributive property.

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

Step 4.1.2.1.2

Multiply $- 1$ by $1$ .

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

Step 4.1.2.1.3

Apply the distributive property.

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

Step 4.1.2.1.4

Multiply $2$ by $1$ .

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

Step 4.1.2.1.5

Apply the distributive property.

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

Step 4.1.2.1.6

Multiply $- 1$ by $1$ .

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

Step 4.1.2.1.7

Apply the distributive property.

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

Step 4.1.2.1.8

Rewrite ${(1 + \sqrt{2})}^{2}$ as $(1 + \sqrt{2}) (1 + \sqrt{2})$ .

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

Step 4.1.2.1.9

Expand $(1 + \sqrt{2}) (1 + \sqrt{2})$ using the FOIL Method.

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Step 4.1.2.1.9.1

Apply the distributive property.

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

Step 4.1.2.1.9.2

Apply the distributive property.

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

Step 4.1.2.1.9.3

Apply the distributive property.

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

Step 4.1.2.1.10

Simplify and combine like terms.

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Step 4.1.2.1.10.1

Simplify each term.

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Step 4.1.2.1.10.1.1

Multiply $1$ by $1$ .

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

Step 4.1.2.1.10.1.2

Multiply $\sqrt{2}$ by $1$ .

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

Step 4.1.2.1.10.1.3

Multiply $\sqrt{2}$ by $1$ .

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

Step 4.1.2.1.10.1.4

Combine using the product rule for radicals.

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

Step 4.1.2.1.10.1.5

Multiply $2$ by $2$ .

$f (1 + \sqrt{2}) = e^{- 1 - \sqrt{2}} + 2 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + (1 + \sqrt{2} + \sqrt{2} + \sqrt{4}) e^{- (1 + \sqrt{2})}$

Step 4.1.2.1.10.1.6

Rewrite $4$ as $2^{2}$ .

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

Step 4.1.2.1.10.1.7

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

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

Step 4.1.2.1.10.2

Add $1$ and $2$ .

$f (1 + \sqrt{2}) = e^{- 1 - \sqrt{2}} + 2 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + (3 + \sqrt{2} + \sqrt{2}) e^{- (1 + \sqrt{2})}$

Step 4.1.2.1.10.3

Add $\sqrt{2}$ and $\sqrt{2}$ .

$f (1 + \sqrt{2}) = e^{- 1 - \sqrt{2}} + 2 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + (3 + 2 \sqrt{2}) e^{- (1 + \sqrt{2})}$

Step 4.1.2.1.11

Apply the distributive property.

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

Step 4.1.2.1.12

Multiply $- 1$ by $1$ .

$f (1 + \sqrt{2}) = e^{- 1 - \sqrt{2}} + 2 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + (3 + 2 \sqrt{2}) e^{- 1 - \sqrt{2}}$

Step 4.1.2.1.13

Apply the distributive property.

$f (1 + \sqrt{2}) = e^{- 1 - \sqrt{2}} + 2 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + 3 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}}$

Step 4.1.2.2

Simplify by adding terms.

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Step 4.1.2.2.1

Add $e^{- 1 - \sqrt{2}}$ and $2 e^{- 1 - \sqrt{2}}$ .

$f (1 + \sqrt{2}) = 3 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}} + 3 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}}$

Step 4.1.2.2.2

Add $3 e^{- 1 - \sqrt{2}}$ and $3 e^{- 1 - \sqrt{2}}$ .

$f (1 + \sqrt{2}) = 2 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}} + 2 \sqrt{2} e^{- 1 - \sqrt{2}}$

Step 4.1.2.2.3

Add $2 \sqrt{2} e^{- 1 - \sqrt{2}}$ and $2 \sqrt{2} e^{- 1 - \sqrt{2}}$ .

$f (1 + \sqrt{2}) = 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}}$

Step 4.1.2.3

The final answer is $4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}}$ .

$4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}}$

Step 4.2

The point found by substituting $1 + \sqrt{2}$ in $f (x) = e^{- x} + 2 x e^{- x} + x^{2} e^{- x}$ is $(1 + \sqrt{2}, 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}})$ . This point can be an inflection point.

$(1 + \sqrt{2}, 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}})$

Step 4.3

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

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Step 4.3.1

Replace the variable $x$ with $1 - \sqrt{2}$ in the expression.

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

Step 4.3.2

Simplify the result.

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Step 4.3.2.1

Simplify each term.

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Step 4.3.2.1.1

Apply the distributive property.

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

Step 4.3.2.1.2

Multiply $- 1$ by $1$ .

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

Step 4.3.2.1.3

Multiply $- - \sqrt{2}$ .

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Step 4.3.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.1.3.2

Multiply $\sqrt{2}$ by $1$ .

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

Step 4.3.2.1.4

Apply the distributive property.

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

Step 4.3.2.1.5

Multiply $2$ by $1$ .

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

Step 4.3.2.1.6

Multiply $- 1$ by $2$ .

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

Step 4.3.2.1.7

Apply the distributive property.

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

Step 4.3.2.1.8

Multiply $- 1$ by $1$ .

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

Step 4.3.2.1.9

Multiply $- - \sqrt{2}$ .

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Step 4.3.2.1.9.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.1.9.2

Multiply $\sqrt{2}$ by $1$ .

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

Step 4.3.2.1.10

Apply the distributive property.

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

Step 4.3.2.1.11

Rewrite ${(1 - \sqrt{2})}^{2}$ as $(1 - \sqrt{2}) (1 - \sqrt{2})$ .

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

Step 4.3.2.1.12

Expand $(1 - \sqrt{2}) (1 - \sqrt{2})$ using the FOIL Method.

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Step 4.3.2.1.12.1

Apply the distributive property.

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

Step 4.3.2.1.12.2

Apply the distributive property.

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

Step 4.3.2.1.12.3

Apply the distributive property.

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

Step 4.3.2.1.13

Simplify and combine like terms.

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Step 4.3.2.1.13.1

Simplify each term.

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Step 4.3.2.1.13.1.1

Multiply $1$ by $1$ .

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

Step 4.3.2.1.13.1.2

Multiply $- \sqrt{2}$ by $1$ .

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

Step 4.3.2.1.13.1.3

Multiply $- 1$ by $1$ .

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

Step 4.3.2.1.13.1.4

Multiply $- \sqrt{2} (- \sqrt{2})$ .

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Step 4.3.2.1.13.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 4.3.2.1.13.1.4.2

Multiply $\sqrt{2}$ by $1$ .

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

Step 4.3.2.1.13.1.4.3

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 4.3.2.1.13.1.4.4

Raise $\sqrt{2}$ to the power of $1$ .

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

Step 4.3.2.1.13.1.4.5

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

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

Step 4.3.2.1.13.1.4.6

Add $1$ and $1$ .

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

Step 4.3.2.1.13.1.5

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 4.3.2.1.13.1.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

Step 4.3.2.1.13.1.5.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.3.2.1.13.1.5.3

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

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

Step 4.3.2.1.13.1.5.4

Cancel the common factor of $2$ .

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Step 4.3.2.1.13.1.5.4.1

Cancel the common factor.

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

Step 4.3.2.1.13.1.5.4.2

Rewrite the expression.

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

Step 4.3.2.1.13.1.5.5

Evaluate the exponent.

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

Step 4.3.2.1.13.2

Add $1$ and $2$ .

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + (3 - \sqrt{2} - \sqrt{2}) e^{- (1 - \sqrt{2})}$

Step 4.3.2.1.13.3

Subtract $\sqrt{2}$ from $- \sqrt{2}$ .

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + (3 - 2 \sqrt{2}) e^{- (1 - \sqrt{2})}$

Step 4.3.2.1.14

Apply the distributive property.

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

Step 4.3.2.1.15

Multiply $- 1$ by $1$ .

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + (3 - 2 \sqrt{2}) e^{- 1 + \sqrt{2}}$

Step 4.3.2.1.16

Multiply $- - \sqrt{2}$ .

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Step 4.3.2.1.16.1

Multiply $- 1$ by $- 1$ .

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + (3 - 2 \sqrt{2}) e^{- 1 + 1 \sqrt{2}}$

Step 4.3.2.1.16.2

Multiply $\sqrt{2}$ by $1$ .

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + (3 - 2 \sqrt{2}) e^{- 1 + \sqrt{2}}$

Step 4.3.2.1.17

Apply the distributive property.

$f (1 - \sqrt{2}) = e^{- 1 + \sqrt{2}} + 2 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + 3 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}}$

Step 4.3.2.2

Simplify by adding terms.

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Step 4.3.2.2.1

Add $e^{- 1 + \sqrt{2}}$ and $2 e^{- 1 + \sqrt{2}}$ .

$f (1 - \sqrt{2}) = 3 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + 3 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}}$

Step 4.3.2.2.2

Add $3 e^{- 1 + \sqrt{2}}$ and $3 e^{- 1 + \sqrt{2}}$ .

$f (1 - \sqrt{2}) = - 2 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}} - 2 \sqrt{2} e^{- 1 + \sqrt{2}}$

Step 4.3.2.2.3

Subtract $2 \sqrt{2} e^{- 1 + \sqrt{2}}$ from $- 2 \sqrt{2} e^{- 1 + \sqrt{2}}$ .

$f (1 - \sqrt{2}) = - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}}$

Step 4.3.2.3

The final answer is $- 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}}$ .

$- 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}}$

Step 4.4

The point found by substituting $1 - \sqrt{2}$ in $f (x) = e^{- x} + 2 x e^{- x} + x^{2} e^{- x}$ is $(1 - \sqrt{2}, - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}})$ . This point can be an inflection point.

$(1 - \sqrt{2}, - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}})$

Step 4.5

Determine the points that could be inflection points.

$(1 + \sqrt{2}, 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}}), (1 - \sqrt{2}, - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}})$

Step 5

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

$(- \infty, 1 - \sqrt{2}) \cup (1 - \sqrt{2}, 1 + \sqrt{2}) \cup (1 + \sqrt{2}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 1 - \sqrt{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 0.51421356$ in the expression.

$f'' (- 0.51421356) = {(- 0.51421356)}^{2} e^{- (- 0.51421356)} - 2 \cdot - 0.51421356 \cdot e^{- (- 0.51421356)} - e^{- (- 0.51421356)}$

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 $- 0.51421356$ to the power of $2$ .

$f'' (- 0.51421356) = 0.26441558 e^{- (- 0.51421356)} - 2 \cdot - 0.51421356 \cdot e^{- (- 0.51421356)} - e^{- (- 0.51421356)}$

Step 6.2.1.2

Multiply $- 1$ by $- 0.51421356$ .

$f'' (- 0.51421356) = 0.26441558 e^{0.51421356} - 2 \cdot - 0.51421356 \cdot e^{- (- 0.51421356)} - e^{- (- 0.51421356)}$

Step 6.2.1.3

Multiply $0.26441558$ by $e^{0.51421356}$ .

$f'' (- 0.51421356) = 0.44218821 - 2 \cdot - 0.51421356 \cdot e^{- (- 0.51421356)} - e^{- (- 0.51421356)}$

Step 6.2.1.4

Multiply $- 2$ by $- 0.51421356$ .

$f'' (- 0.51421356) = 0.44218821 + 1.02842712 \cdot e^{- (- 0.51421356)} - e^{- (- 0.51421356)}$

Step 6.2.1.5

Multiply $- 1$ by $- 0.51421356$ .

$f'' (- 0.51421356) = 0.44218821 + 1.02842712 \cdot e^{0.51421356} - e^{- (- 0.51421356)}$

Step 6.2.1.6

Multiply $1.02842712$ by $e^{0.51421356}$ .

$f'' (- 0.51421356) = 0.44218821 + 1.71986213 - e^{- (- 0.51421356)}$

Step 6.2.1.7

Multiply $- 1$ by $- 0.51421356$ .

$f'' (- 0.51421356) = 0.44218821 + 1.71986213 - e^{0.51421356}$

Step 6.2.2

Simplify by adding terms.

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Step 6.2.2.1

Add $0.44218821$ and $1.71986213$ .

$f'' (- 0.51421356) = 2.16205035 - e^{0.51421356}$

Step 6.2.2.2

Subtract $e^{0.51421356}$ from $2.16205035$ .

$f'' (- 0.51421356) = 0.48972754$

Step 6.2.3

The final answer is $0.48972754$ .

$0.48972754$

Step 6.3

At $- 0.51421356$ , the second derivative is $0.48972754$ . Since this is positive, the second derivative is increasing on the interval $(- \infty, 1 - \sqrt{2})$ .

Increasing on $(- \infty, 1 - \sqrt{2})$ since $f'' (x) > 0$

Step 7

Substitute a value from the interval $(1 - \sqrt{2}, 1 + \sqrt{2})$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

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

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

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

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

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

Step 7.2.1.2

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

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

Step 7.2.1.3

Multiply $- 1$ by $1$ .

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

Step 7.2.1.4

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

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

Step 7.2.1.5

Multiply $- 2$ by $1$ .

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

Step 7.2.1.6

Multiply $- 1$ by $1$ .

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

Step 7.2.1.7

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

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

Step 7.2.1.8

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

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

Step 7.2.1.9

Move the negative in front of the fraction.

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

Step 7.2.1.10

Replace $e$ with an approximation.

$f'' (1) = \frac{1}{e^{1}} - \frac{2}{{2.71828182}^{1}} - e^{- (1)}$

Step 7.2.1.11

Raise $2.71828182$ to the power of $1$ .

$f'' (1) = \frac{1}{e^{1}} - \frac{2}{2.71828182} - e^{- (1)}$

Step 7.2.1.12

Divide $2$ by $2.71828182$ .

$f'' (1) = \frac{1}{e^{1}} - 1 \cdot 0.73575888 - e^{- (1)}$

Step 7.2.1.13

Multiply $- 1$ by $0.73575888$ .

$f'' (1) = \frac{1}{e^{1}} - 0.73575888 - e^{- (1)}$

Step 7.2.1.14

Multiply $- 1$ by $1$ .

$f'' (1) = \frac{1}{e^{1}} - 0.73575888 - e^{- 1}$

Step 7.2.1.15

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

$f'' (1) = \frac{1}{e^{1}} - 0.73575888 - \frac{1}{e^{1}}$

Step 7.2.2

Combine fractions.

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Step 7.2.2.1

Combine the numerators over the common denominator.

$f'' (1) = - 0.73575888 + \frac{1 - 1}{e^{1}}$

Step 7.2.2.2

Simplify the expression.

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Step 7.2.2.2.1

Subtract $1$ from $1$ .

$f'' (1) = - 0.73575888 + \frac{0}{e^{1}}$

Step 7.2.2.2.2

Divide $0$ by $e^{1}$ .

$f'' (1) = - 0.73575888 + 0$

Step 7.2.2.2.3

Add $- 0.73575888$ and $0$ .

$f'' (1) = - 0.73575888$

Step 7.2.3

The final answer is $- 0.73575888$ .

$- 0.73575888$

Step 7.3

At $1$ , the second derivative is $- 0.73575888$ . Since this is negative, the second derivative is decreasing on the interval $(1 - \sqrt{2}, 1 + \sqrt{2})$

Decreasing on $(1 - \sqrt{2}, 1 + \sqrt{2})$ since $f'' (x) < 0$

Step 8

Substitute a value from the interval $(1 + \sqrt{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $2.51421356$ in the expression.

$f'' (2.51421356) = {(2.51421356)}^{2} e^{- (2.51421356)} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Raise $2.51421356$ to the power of $2$ .

$f'' (2.51421356) = 6.32126983 e^{- (2.51421356)} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.2

Multiply $- 1$ by $2.51421356$ .

$f'' (2.51421356) = 6.32126983 e^{- 2.51421356} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.3

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

$f'' (2.51421356) = 6.32126983 (\frac{1}{e^{2.51421356}}) - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.4

Combine $6.32126983$ and $\frac{1}{e^{2.51421356}}$ .

$f'' (2.51421356) = \frac{6.32126983}{e^{2.51421356}} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.5

Replace $e$ with an approximation.

$f'' (2.51421356) = \frac{6.32126983}{{2.71828182}^{2.51421356}} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.6

Raise $2.71828182$ to the power of $2.51421356$ .

$f'' (2.51421356) = \frac{6.32126983}{12.35688703} - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.7

Divide $6.32126983$ by $12.35688703$ .

$f'' (2.51421356) = 0.51155843 - 2 \cdot 2.51421356 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.8

Multiply $- 2$ by $2.51421356$ .

$f'' (2.51421356) = 0.51155843 - 5.02842712 \cdot e^{- (2.51421356)} - e^{- (2.51421356)}$

Step 8.2.1.9

Multiply $- 1$ by $2.51421356$ .

$f'' (2.51421356) = 0.51155843 - 5.02842712 \cdot e^{- 2.51421356} - e^{- (2.51421356)}$

Step 8.2.1.10

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

$f'' (2.51421356) = 0.51155843 - 5.02842712 \cdot \frac{1}{e^{2.51421356}} - e^{- (2.51421356)}$

Step 8.2.1.11

Combine $- 5.02842712$ and $\frac{1}{e^{2.51421356}}$ .

$f'' (2.51421356) = 0.51155843 + \frac{- 5.02842712}{e^{2.51421356}} - e^{- (2.51421356)}$

Step 8.2.1.12

Move the negative in front of the fraction.

$f'' (2.51421356) = 0.51155843 - \frac{5.02842712}{e^{2.51421356}} - e^{- (2.51421356)}$

Step 8.2.1.13

Replace $e$ with an approximation.

$f'' (2.51421356) = 0.51155843 - \frac{5.02842712}{{2.71828182}^{2.51421356}} - e^{- (2.51421356)}$

Step 8.2.1.14

Raise $2.71828182$ to the power of $2.51421356$ .

$f'' (2.51421356) = 0.51155843 - \frac{5.02842712}{12.35688703} - e^{- (2.51421356)}$

Step 8.2.1.15

Divide $5.02842712$ by $12.35688703$ .

$f'' (2.51421356) = 0.51155843 - 1 \cdot 0.40693316 - e^{- (2.51421356)}$

Step 8.2.1.16

Multiply $- 1$ by $0.40693316$ .

$f'' (2.51421356) = 0.51155843 - 0.40693316 - e^{- (2.51421356)}$

Step 8.2.1.17

Multiply $- 1$ by $2.51421356$ .

$f'' (2.51421356) = 0.51155843 - 0.40693316 - e^{- 2.51421356}$

Step 8.2.1.18

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

$f'' (2.51421356) = 0.51155843 - 0.40693316 - \frac{1}{e^{2.51421356}}$

Step 8.2.2

Simplify by adding terms.

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Step 8.2.2.1

Subtract $0.40693316$ from $0.51155843$ .

$f'' (2.51421356) = 0.10462527 - \frac{1}{e^{2.51421356}}$

Step 8.2.2.2

Subtract $\frac{1}{e^{2.51421356}}$ from $0.10462527$ .

$f'' (2.51421356) = 0.02369874$

Step 8.2.3

The final answer is $0.02369874$ .

$0.02369874$

Step 8.3

At $2.51421356$ , the second derivative is $0.02369874$ . Since this is positive, the second derivative is increasing on the interval $(1 + \sqrt{2}, \infty)$ .

Increasing on $(1 + \sqrt{2}, \infty)$ 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 $(1 - \sqrt{2}, - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}}), (1 + \sqrt{2}, 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}})$ .

$(1 - \sqrt{2}, - 4 \sqrt{2} e^{- 1 + \sqrt{2}} + 6 e^{- 1 + \sqrt{2}}), (1 + \sqrt{2}, 4 \sqrt{2} e^{- 1 - \sqrt{2}} + 6 e^{- 1 - \sqrt{2}})$

Step 10