Calculus Examples

$f (x) = 10 {(x - 1)}^{2} \cdot e^{- x}$

Step 1

Find the first derivative of the function.

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Step 1.1

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

$\frac{d}{d x} [10 ((x - 1) (x - 1)) \cdot e^{- x}]$

Step 1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 1.2.1

Apply the distributive property.

$\frac{d}{d x} [10 (x (x - 1) - 1 (x - 1)) \cdot e^{- x}]$

Step 1.2.2

Apply the distributive property.

$\frac{d}{d x} [10 (x \cdot x + x \cdot - 1 - 1 (x - 1)) \cdot e^{- x}]$

Step 1.2.3

Apply the distributive property.

$\frac{d}{d x} [10 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 1.3

Simplify and combine like terms.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Multiply $x$ by $x$ .

$\frac{d}{d x} [10 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 1.3.1.2

Move $- 1$ to the left of $x$ .

$\frac{d}{d x} [10 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\frac{d}{d x} [10 (x^{2} - x - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 1.3.1.4

Rewrite $- 1 x$ as $- x$ .

$\frac{d}{d x} [10 (x^{2} - x - x - 1 \cdot - 1) \cdot e^{- x}]$

Step 1.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [10 (x^{2} - x - x + 1) \cdot e^{- x}]$

Step 1.3.2

Subtract $x$ from $- x$ .

$\frac{d}{d x} [10 (x^{2} - 2 x + 1) \cdot e^{- x}]$

Step 1.4

Since $10$ is constant with respect to $x$ , the derivative of $10 (x^{2} - 2 x + 1) e^{- x}$ with respect to $x$ is $10 \frac{d}{d x} [(x^{2} - 2 x + 1) e^{- x}]$ .

$10 \frac{d}{d x} [(x^{2} - 2 x + 1) e^{- x}]$

Step 1.5

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 $f (x) = x^{2} - 2 x + 1$ and $g (x) = e^{- x}$ .

$10 ((x^{2} - 2 x + 1) \frac{d}{d x} [e^{- x}] + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.6

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 1.6.1

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

$10 ((x^{2} - 2 x + 1) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.6.2

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

$10 ((x^{2} - 2 x + 1) (e^{u} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.6.3

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

$10 ((x^{2} - 2 x + 1) (e^{- x} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7

Differentiate.

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Step 1.7.1

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

$10 ((x^{2} - 2 x + 1) e^{- x} (- \frac{d}{d x} [x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7.2

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

$10 ((x^{2} - 2 x + 1) e^{- x} (- 1 \cdot 1) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7.3

Simplify the expression.

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Step 1.7.3.1

Multiply $- 1$ by $1$ .

$10 ((x^{2} - 2 x + 1) e^{- x} \cdot - 1 + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7.3.2

Move $- 1$ to the left of $(x^{2} - 2 x + 1) e^{- x}$ .

$10 (- 1 \cdot ((x^{2} - 2 x + 1) e^{- x}) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7.3.3

Rewrite $- 1 (x^{2} - 2 x + 1)$ as $- (x^{2} - 2 x + 1)$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 1.7.4

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]))$

Step 1.7.5

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]))$

Step 1.7.6

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 1.7.7

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 \cdot 1 + \frac{d}{d x} [1]))$

Step 1.7.8

Multiply $- 2$ by $1$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 + \frac{d}{d x} [1]))$

Step 1.7.9

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 + 0))$

Step 1.7.10

Add $2 x - 2$ and $0$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2))$

Step 1.8

Simplify.

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Step 1.8.1

Apply the distributive property.

$10 ((- x^{2} - (- 2 x) - 1 \cdot 1) e^{- x} + e^{- x} (2 x - 2))$

Step 1.8.2

Apply the distributive property.

$10 (- x^{2} e^{- x} - (- 2 x) e^{- x} - 1 \cdot 1 e^{- x} + e^{- x} (2 x - 2))$

Step 1.8.3

Apply the distributive property.

$10 (- x^{2} e^{- x} - (- 2 x) e^{- x} - 1 \cdot 1 e^{- x} + e^{- x} (2 x) + e^{- x} \cdot - 2)$

Step 1.8.4

Apply the distributive property.

$10 (- x^{2} e^{- x}) + 10 (- (- 2 x) e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5

Combine terms.

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Step 1.8.5.1

Multiply $- 1$ by $10$ .

$- 10 (x^{2} e^{- x}) + 10 (- (- 2 x) e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.2

Multiply $- 2$ by $- 1$ .

$- 10 x^{2} e^{- x} + 10 (2 x e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.3

Multiply $2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 (x e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.4

Multiply $- 1$ by $1$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 10 (- 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.5

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

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 10 (- e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.6

Multiply $- 1$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.7

Multiply $2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 (e^{- x} (x)) + 10 (e^{- x} \cdot - 2)$

Step 1.8.5.8

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

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 e^{- x} x + 10 (- 2 \cdot e^{- x})$

Step 1.8.5.9

Multiply $- 2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 e^{- x} x - 20 e^{- x}$

Step 1.8.5.10

Add $20 x e^{- x}$ and $20 e^{- x} x$ .

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Step 1.8.5.10.1

Move $e^{- x}$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 20 x e^{- x} - 10 e^{- x} - 20 e^{- x}$

Step 1.8.5.10.2

Add $20 x e^{- x}$ and $20 x e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 10 e^{- x} - 20 e^{- x}$

Step 1.8.5.11

Subtract $20 e^{- x}$ from $- 10 e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$

Step 1.8.6

Reorder terms.

$- 10 e^{- x} x^{2} + 40 e^{- x} x - 30 e^{- x}$

Step 1.8.7

Reorder factors in $- 10 e^{- x} x^{2} + 40 e^{- x} x - 30 e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [- 10 x^{2} e^{- x}] + \frac{d}{d x} [40 x e^{- x}] + \frac{d}{d x} [- 30 e^{- x}]$ .

$f'' (x) = \frac{d}{d x} (- 10 x^{2} e^{- x}) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2

Evaluate $\frac{d}{d x} [- 10 x^{2} e^{- x}]$ .

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Step 2.2.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x^{2} e^{- x}$ with respect to $x$ is $- 10 \frac{d}{d x} [x^{2} e^{- x}]$ .

$f'' (x) = - 10 \frac{d}{d x} (x^{2} e^{- x}) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

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

$f'' (x) = - 10 (x^{2} \frac{d}{d x} (e^{- x}) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

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 $g (x) = - x$ .

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Step 2.2.3.1

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

$f'' (x) = - 10 (x^{2} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

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 $a$ = $e$ .

$f'' (x) = - 10 (x^{2} (e^{u_{1}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.3.3

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

$f'' (x) = - 10 (x^{2} (e^{- x} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.4

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

$f'' (x) = - 10 (x^{2} (e^{- x} (- \frac{d}{d x} x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.5

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

$f'' (x) = - 10 (x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.6

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

$f'' (x) = - 10 (x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} (2 x)) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.7

Multiply $- 1$ by $1$ .

$f'' (x) = - 10 (x^{2} (e^{- x} \cdot - 1) + e^{- x} (2 x)) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.8

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

$f'' (x) = - 10 (x^{2} (- 1 \cdot e^{- x}) + e^{- x} (2 x)) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.2.9

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + \frac{d}{d x} (40 x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3

Evaluate $\frac{d}{d x} [40 x e^{- x}]$ .

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Step 2.3.1

Since $40$ is constant with respect to $x$ , the derivative of $40 x e^{- x}$ with respect to $x$ is $40 \frac{d}{d x} [x e^{- x}]$ .

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 \frac{d}{d x} (x e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.2

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x \frac{d}{d x} (e^{- x}) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.3

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 2.3.3.1

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.3.2

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{u_{2}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.3.3

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{- x} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.4

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{- x} (- \frac{d}{d x} x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.5

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.6

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{- x} (- 1 \cdot 1)) + e^{- x} \cdot 1) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.7

Multiply $- 1$ by $1$ .

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (e^{- x} \cdot - 1) + e^{- x} \cdot 1) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.8

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- 1 \cdot e^{- x}) + e^{- x} \cdot 1) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.9

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x} \cdot 1) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.3.10

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) + \frac{d}{d x} (- 30 e^{- x})$

Step 2.4

Evaluate $\frac{d}{d x} [- 30 e^{- x}]$ .

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Step 2.4.1

Since $- 30$ is constant with respect to $x$ , the derivative of $- 30 e^{- x}$ with respect to $x$ is $- 30 \frac{d}{d x} [e^{- x}]$ .

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 \frac{d}{d x} e^{- x}$

Step 2.4.2

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 2.4.2.1

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (\frac{d}{d u (3)} (e^{u_{3}}) \frac{d}{d x} (- x))$

Step 2.4.2.2

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (e^{u_{3}} \frac{d}{d x} (- x))$

Step 2.4.2.3

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (e^{- x} \frac{d}{d x} (- x))$

Step 2.4.3

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (e^{- x} (- \frac{d}{d x} x))$

Step 2.4.4

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (e^{- x} (- 1 \cdot 1))$

Step 2.4.5

Multiply $- 1$ by $1$ .

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (e^{- x} \cdot - 1)$

Step 2.4.6

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (- 1 \cdot e^{- x})$

Step 2.4.7

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

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) - 30 (- e^{- x})$

Step 2.4.8

Multiply $- 1$ by $- 30$ .

$f'' (x) = - 10 (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) + 30 e^{- x}$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$f'' (x) = - 10 (x^{2} (- e^{- x})) - 10 (e^{- x} (2 x)) + 40 (x (- e^{- x}) + e^{- x}) + 30 e^{- x}$

Step 2.5.2

Apply the distributive property.

$f'' (x) = - 10 (x^{2} (- e^{- x})) - 10 (e^{- x} (2 x)) + 40 (x (- e^{- x})) + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3

Combine terms.

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Step 2.5.3.1

Multiply $- 1$ by $- 10$ .

$f'' (x) = 10 (x^{2} (e^{- x})) - 10 (e^{- x} (2 x)) + 40 (x (- e^{- x})) + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3.2

Multiply $2$ by $- 10$ .

$f'' (x) = 10 x^{2} e^{- x} - 20 (e^{- x} (x)) + 40 (x (- e^{- x})) + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3.3

Multiply $- 1$ by $40$ .

$f'' (x) = 10 x^{2} e^{- x} - 20 e^{- x} x - 40 (x (e^{- x})) + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3.4

Subtract $40 x e^{- x}$ from $- 20 e^{- x} x$ .

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Step 2.5.3.4.1

Move $e^{- x}$ .

$f'' (x) = 10 x^{2} e^{- x} - 20 x e^{- x} - 40 x e^{- x} + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3.4.2

Subtract $40 x e^{- x}$ from $- 20 x e^{- x}$ .

$f'' (x) = 10 x^{2} e^{- x} - 60 x e^{- x} + 40 e^{- x} + 30 e^{- x}$

Step 2.5.3.5

Add $40 e^{- x}$ and $30 e^{- x}$ .

$f'' (x) = 10 x^{2} e^{- x} - 60 x e^{- x} + 70 e^{- x}$

Step 2.5.4

Reorder terms.

$f'' (x) = 10 e^{- x} x^{2} - 60 e^{- x} x + 70 e^{- x}$

Step 2.5.5

Reorder factors in $10 e^{- x} x^{2} - 60 e^{- x} x + 70 e^{- x}$ .

$f'' (x) = 10 x^{2} e^{- x} - 60 x e^{- x} + 70 e^{- x}$

Step 3

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

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x} = 0$

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

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

$\frac{d}{d x} [10 ((x - 1) (x - 1)) \cdot e^{- x}]$

Step 4.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 4.1.2.1

Apply the distributive property.

$\frac{d}{d x} [10 (x (x - 1) - 1 (x - 1)) \cdot e^{- x}]$

Step 4.1.2.2

Apply the distributive property.

$\frac{d}{d x} [10 (x \cdot x + x \cdot - 1 - 1 (x - 1)) \cdot e^{- x}]$

Step 4.1.2.3

Apply the distributive property.

$\frac{d}{d x} [10 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 4.1.3

Simplify and combine like terms.

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Step 4.1.3.1

Simplify each term.

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Step 4.1.3.1.1

Multiply $x$ by $x$ .

$\frac{d}{d x} [10 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 4.1.3.1.2

Move $- 1$ to the left of $x$ .

$\frac{d}{d x} [10 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 4.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

$\frac{d}{d x} [10 (x^{2} - x - 1 x - 1 \cdot - 1) \cdot e^{- x}]$

Step 4.1.3.1.4

Rewrite $- 1 x$ as $- x$ .

$\frac{d}{d x} [10 (x^{2} - x - x - 1 \cdot - 1) \cdot e^{- x}]$

Step 4.1.3.1.5

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [10 (x^{2} - x - x + 1) \cdot e^{- x}]$

Step 4.1.3.2

Subtract $x$ from $- x$ .

$\frac{d}{d x} [10 (x^{2} - 2 x + 1) \cdot e^{- x}]$

Step 4.1.4

Since $10$ is constant with respect to $x$ , the derivative of $10 (x^{2} - 2 x + 1) e^{- x}$ with respect to $x$ is $10 \frac{d}{d x} [(x^{2} - 2 x + 1) e^{- x}]$ .

$10 \frac{d}{d x} [(x^{2} - 2 x + 1) e^{- x}]$

Step 4.1.5

$10 ((x^{2} - 2 x + 1) \frac{d}{d x} [e^{- x}] + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.6

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 4.1.6.1

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

$10 ((x^{2} - 2 x + 1) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.6.2

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

$10 ((x^{2} - 2 x + 1) (e^{u} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.6.3

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

$10 ((x^{2} - 2 x + 1) (e^{- x} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7

Differentiate.

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Step 4.1.7.1

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

$10 ((x^{2} - 2 x + 1) e^{- x} (- \frac{d}{d x} [x]) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7.2

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

$10 ((x^{2} - 2 x + 1) e^{- x} (- 1 \cdot 1) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7.3

Simplify the expression.

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Step 4.1.7.3.1

Multiply $- 1$ by $1$ .

$10 ((x^{2} - 2 x + 1) e^{- x} \cdot - 1 + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7.3.2

Move $- 1$ to the left of $(x^{2} - 2 x + 1) e^{- x}$ .

$10 (- 1 \cdot ((x^{2} - 2 x + 1) e^{- x}) + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7.3.3

Rewrite $- 1 (x^{2} - 2 x + 1)$ as $- (x^{2} - 2 x + 1)$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} \frac{d}{d x} [x^{2} - 2 x + 1])$

Step 4.1.7.4

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]))$

Step 4.1.7.5

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]))$

Step 4.1.7.6

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 4.1.7.7

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

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 \cdot 1 + \frac{d}{d x} [1]))$

Step 4.1.7.8

Multiply $- 2$ by $1$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 + \frac{d}{d x} [1]))$

Step 4.1.7.9

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2 + 0))$

Step 4.1.7.10

Add $2 x - 2$ and $0$ .

$10 (- (x^{2} - 2 x + 1) e^{- x} + e^{- x} (2 x - 2))$

Step 4.1.8

Simplify.

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Step 4.1.8.1

Apply the distributive property.

$10 ((- x^{2} - (- 2 x) - 1 \cdot 1) e^{- x} + e^{- x} (2 x - 2))$

Step 4.1.8.2

Apply the distributive property.

$10 (- x^{2} e^{- x} - (- 2 x) e^{- x} - 1 \cdot 1 e^{- x} + e^{- x} (2 x - 2))$

Step 4.1.8.3

Apply the distributive property.

$10 (- x^{2} e^{- x} - (- 2 x) e^{- x} - 1 \cdot 1 e^{- x} + e^{- x} (2 x) + e^{- x} \cdot - 2)$

Step 4.1.8.4

Apply the distributive property.

$10 (- x^{2} e^{- x}) + 10 (- (- 2 x) e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5

Combine terms.

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Step 4.1.8.5.1

Multiply $- 1$ by $10$ .

$- 10 (x^{2} e^{- x}) + 10 (- (- 2 x) e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.2

Multiply $- 2$ by $- 1$ .

$- 10 x^{2} e^{- x} + 10 (2 x e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.3

Multiply $2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 (x e^{- x}) + 10 (- 1 \cdot 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.4

Multiply $- 1$ by $1$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 10 (- 1 e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.5

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

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 10 (- e^{- x}) + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.6

Multiply $- 1$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 10 (e^{- x} (2 x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.7

Multiply $2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 (e^{- x} (x)) + 10 (e^{- x} \cdot - 2)$

Step 4.1.8.5.8

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

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 e^{- x} x + 10 (- 2 \cdot e^{- x})$

Step 4.1.8.5.9

Multiply $- 2$ by $10$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} - 10 e^{- x} + 20 e^{- x} x - 20 e^{- x}$

Step 4.1.8.5.10

Add $20 x e^{- x}$ and $20 e^{- x} x$ .

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Step 4.1.8.5.10.1

Move $e^{- x}$ .

$- 10 x^{2} e^{- x} + 20 x e^{- x} + 20 x e^{- x} - 10 e^{- x} - 20 e^{- x}$

Step 4.1.8.5.10.2

Add $20 x e^{- x}$ and $20 x e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 10 e^{- x} - 20 e^{- x}$

Step 4.1.8.5.11

Subtract $20 e^{- x}$ from $- 10 e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$

Step 4.1.8.6

Reorder terms.

$- 10 e^{- x} x^{2} + 40 e^{- x} x - 30 e^{- x}$

Step 4.1.8.7

Reorder factors in $- 10 e^{- x} x^{2} + 40 e^{- x} x - 30 e^{- x}$ .

$f' (x) = - 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x} = 0$

Step 5.2

Factor the left side of the equation.

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Step 5.2.1

Factor $10 e^{- x}$ out of $- 10 x^{2} e^{- x} + 40 x e^{- x} - 30 e^{- x}$ .

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Step 5.2.1.1

Factor $10 e^{- x}$ out of $- 10 x^{2} e^{- x}$ .

$10 e^{- x} (- x^{2}) + 40 x e^{- x} - 30 e^{- x} = 0$

Step 5.2.1.2

Factor $10 e^{- x}$ out of $40 x e^{- x}$ .

$10 e^{- x} (- x^{2}) + 10 e^{- x} (4 x) - 30 e^{- x} = 0$

Step 5.2.1.3

Factor $10 e^{- x}$ out of $- 30 e^{- x}$ .

$10 e^{- x} (- x^{2}) + 10 e^{- x} (4 x) + 10 e^{- x} (- 3) = 0$

Step 5.2.1.4

Factor $10 e^{- x}$ out of $10 e^{- x} (- x^{2}) + 10 e^{- x} (4 x)$ .

$10 e^{- x} (- x^{2} + 4 x) + 10 e^{- x} (- 3) = 0$

Step 5.2.1.5

Factor $10 e^{- x}$ out of $10 e^{- x} (- x^{2} + 4 x) + 10 e^{- x} (- 3)$ .

$10 e^{- x} (- x^{2} + 4 x - 3) = 0$

Step 5.2.2

Factor.

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Step 5.2.2.1

Factor by grouping.

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Step 5.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 3 = 3$ and whose sum is $b = 4$ .

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Step 5.2.2.1.1.1

Factor $4$ out of $4 x$ .

$10 e^{- x} (- x^{2} + 4 (x) - 3) = 0$

Step 5.2.2.1.1.2

Rewrite $4$ as $1$ plus $3$

$10 e^{- x} (- x^{2} + (1 + 3) x - 3) = 0$

Step 5.2.2.1.1.3

Apply the distributive property.

$10 e^{- x} (- x^{2} + 1 x + 3 x - 3) = 0$

Step 5.2.2.1.1.4

Multiply $x$ by $1$ .

$10 e^{- x} (- x^{2} + x + 3 x - 3) = 0$

Step 5.2.2.1.2

Factor out the greatest common factor from each group.

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Step 5.2.2.1.2.1

Group the first two terms and the last two terms.

$10 e^{- x} ((- x^{2} + x) + 3 x - 3) = 0$

Step 5.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$10 e^{- x} (x (- x + 1) - 3 (- x + 1)) = 0$

Step 5.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $- x + 1$ .

$10 e^{- x} ((- x + 1) (x - 3)) = 0$

Step 5.2.2.2

Remove unnecessary parentheses.

$10 e^{- x} (- x + 1) (x - 3) = 0$

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 $0$ .

$e^{- x} = 0$

$- x + 1 = 0$

$x - 3 = 0$

Step 5.4

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

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Step 5.4.1

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

$e^{- x} = 0$

Step 5.4.2

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

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

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 $e^{- x} = 0$

No solution

Step 5.5

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

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Step 5.5.1

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

$- x + 1 = 0$

Step 5.5.2

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

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Step 5.5.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 5.5.2.2

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

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Step 5.5.2.2.1

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

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

Step 5.5.2.2.2

Simplify the left side.

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Step 5.5.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.5.2.2.2.2

Divide $x$ by $1$ .

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

Step 5.5.2.2.3

Simplify the right side.

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Step 5.5.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 5.6

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 5.6.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 5.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 5.7

The final solution is all the values that make $10 e^{- x} (- x + 1) (x - 3) = 0$ true.

$x = 1, 3$

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.

$x = 1, 3$

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.

$10 {(1)}^{2} e^{- (1)} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

One to any power is one.

$10 \cdot 1 e^{- (1)} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.2

Multiply $10$ by $1$ .

$10 e^{- (1)} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.3

Multiply $- 1$ by $1$ .

$10 e^{- 1} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.4

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

$10 \frac{1}{e} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.5

Combine $10$ and $\frac{1}{e}$ .

$\frac{10}{e} - 60 \cdot 1 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.6

Multiply $- 60$ by $1$ .

$\frac{10}{e} - 60 e^{- (1)} + 70 e^{- (1)}$

Step 9.1.7

Multiply $- 1$ by $1$ .

$\frac{10}{e} - 60 e^{- 1} + 70 e^{- (1)}$

Step 9.1.8

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

$\frac{10}{e} - 60 \frac{1}{e} + 70 e^{- (1)}$

Step 9.1.9

Combine $- 60$ and $\frac{1}{e}$ .

$\frac{10}{e} + \frac{- 60}{e} + 70 e^{- (1)}$

Step 9.1.10

Move the negative in front of the fraction.

$\frac{10}{e} - \frac{60}{e} + 70 e^{- (1)}$

Step 9.1.11

Multiply $- 1$ by $1$ .

$\frac{10}{e} - \frac{60}{e} + 70 e^{- 1}$

Step 9.1.12

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

$\frac{10}{e} - \frac{60}{e} + 70 \frac{1}{e}$

Step 9.1.13

Combine $70$ and $\frac{1}{e}$ .

$\frac{10}{e} - \frac{60}{e} + \frac{70}{e}$

Step 9.2

Combine fractions.

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Step 9.2.1

Combine the numerators over the common denominator.

$\frac{10 - 60 + 70}{e}$

Step 9.2.2

Simplify by adding and subtracting.

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Step 9.2.2.1

Subtract $60$ from $10$ .

$\frac{- 50 + 70}{e}$

Step 9.2.2.2

Add $- 50$ and $70$ .

$\frac{20}{e}$

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.

$f (1) = 10 {((1) - 1)}^{2} \cdot e^{- (1)}$

Step 11.2

Simplify the result.

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Step 11.2.1

Subtract $1$ from $1$ .

$f (1) = 10 \cdot 0^{2} \cdot e^{- (1)}$

Step 11.2.2

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

$f (1) = 10 \cdot 0 \cdot e^{- (1)}$

Step 11.2.3

Multiply $10$ by $0$ .

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

Step 11.2.4

Multiply $- 1$ by $1$ .

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

Step 11.2.5

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

$f (1) = 0 \cdot \frac{1}{e}$

Step 11.2.6

Multiply $0$ by $\frac{1}{e}$ .

$f (1) = 0$

Step 11.2.7

The final answer is $0$ .

$y = 0$

Step 12

Evaluate the second derivative at $x = 3$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$10 {(3)}^{2} e^{- (3)} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

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

$10 \cdot 9 e^{- (3)} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.2

Multiply $10$ by $9$ .

$90 e^{- (3)} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.3

Multiply $- 1$ by $3$ .

$90 e^{- 3} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.4

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

$90 \frac{1}{e^{3}} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.5

Combine $90$ and $\frac{1}{e^{3}}$ .

$\frac{90}{e^{3}} - 60 \cdot 3 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.6

Multiply $- 60$ by $3$ .

$\frac{90}{e^{3}} - 180 e^{- (3)} + 70 e^{- (3)}$

Step 13.1.7

Multiply $- 1$ by $3$ .

$\frac{90}{e^{3}} - 180 e^{- 3} + 70 e^{- (3)}$

Step 13.1.8

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

$\frac{90}{e^{3}} - 180 \frac{1}{e^{3}} + 70 e^{- (3)}$

Step 13.1.9

Combine $- 180$ and $\frac{1}{e^{3}}$ .

$\frac{90}{e^{3}} + \frac{- 180}{e^{3}} + 70 e^{- (3)}$

Step 13.1.10

Move the negative in front of the fraction.

$\frac{90}{e^{3}} - \frac{180}{e^{3}} + 70 e^{- (3)}$

Step 13.1.11

Multiply $- 1$ by $3$ .

$\frac{90}{e^{3}} - \frac{180}{e^{3}} + 70 e^{- 3}$

Step 13.1.12

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

$\frac{90}{e^{3}} - \frac{180}{e^{3}} + 70 \frac{1}{e^{3}}$

Step 13.1.13

Combine $70$ and $\frac{1}{e^{3}}$ .

$\frac{90}{e^{3}} - \frac{180}{e^{3}} + \frac{70}{e^{3}}$

Step 13.2

Combine fractions.

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Step 13.2.1

Combine the numerators over the common denominator.

$\frac{90 - 180 + 70}{e^{3}}$

Step 13.2.2

Simplify the expression.

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Step 13.2.2.1

Subtract $180$ from $90$ .

$\frac{- 90 + 70}{e^{3}}$

Step 13.2.2.2

Add $- 90$ and $70$ .

$\frac{- 20}{e^{3}}$

Step 13.2.2.3

Move the negative in front of the fraction.

$- \frac{20}{e^{3}}$

Step 14

$x = 3$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 3$ is a local maximum

Step 15

Find the y-value when $x = 3$ .

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Step 15.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = 10 {((3) - 1)}^{2} \cdot e^{- (3)}$

Step 15.2

Simplify the result.

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Step 15.2.1

Subtract $1$ from $3$ .

$f (3) = 10 \cdot 2^{2} \cdot e^{- (3)}$

Step 15.2.2

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

$f (3) = 10 \cdot 4 \cdot e^{- (3)}$

Step 15.2.3

Multiply $10$ by $4$ .

$f (3) = 40 \cdot e^{- (3)}$

Step 15.2.4

Multiply $- 1$ by $3$ .

$f (3) = 40 \cdot e^{- 3}$

Step 15.2.5

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

$f (3) = 40 \cdot \frac{1}{e^{3}}$

Step 15.2.6

Combine $40$ and $\frac{1}{e^{3}}$ .

$f (3) = \frac{40}{e^{3}}$

Step 15.2.7

The final answer is $\frac{40}{e^{3}}$ .

$y = \frac{40}{e^{3}}$

Step 16

These are the local extrema for $f (x) = 10 {(x - 1)}^{2} \cdot e^{- x}$ .

$(1, 0)$ is a local minima

$(3, \frac{40}{e^{3}})$ is a local maxima

Step 17