Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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 $g (x) = x^{3} - 3 x$ .

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Step 1.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{3} - 3 x]$

Step 1.1.2

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

$e^{u} \frac{d}{d x} [x^{3} - 3 x]$

Step 1.1.3

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

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

Step 1.2

Differentiate.

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Step 1.2.1

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

$e^{x^{3} - 3 x} (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x])$

Step 1.2.2

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

$e^{x^{3} - 3 x} (3 x^{2} + \frac{d}{d x} [- 3 x])$

Step 1.2.3

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

$e^{x^{3} - 3 x} (3 x^{2} - 3 \frac{d}{d x} [x])$

Step 1.2.4

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

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

Step 1.2.5

Multiply $- 3$ by $1$ .

$e^{x^{3} - 3 x} (3 x^{2} - 3)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{x^{3} - 3 x}$ and $g (x) = 3 x^{2} - 3$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $3 x^{2} - 3$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 3]$ .

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

Step 2.2.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $3$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x + \frac{d}{d x} (- 3)) + (3 x^{2} - 3) \frac{d}{d x} (e^{x^{3} - 3 x})$

Step 2.2.5

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x + 0) + (3 x^{2} - 3) \frac{d}{d x} (e^{x^{3} - 3 x})$

Step 2.2.6

Add $6 x$ and $0$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) \frac{d}{d x} (e^{x^{3} - 3 x})$

Step 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 $g (x) = x^{3} - 3 x$ .

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Step 2.3.1

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) (\frac{d}{d u} (e^{u}) \frac{d}{d x} (x^{3} - 3 x))$

Step 2.3.2

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) (e^{u} \frac{d}{d x} (x^{3} - 3 x))$

Step 2.3.3

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) (e^{x^{3} - 3 x} \frac{d}{d x} (x^{3} - 3 x))$

Step 2.4

Differentiate.

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Step 2.4.1

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) e^{x^{3} - 3 x} (\frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 3 x))$

Step 2.4.2

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) e^{x^{3} - 3 x} (3 x^{2} + \frac{d}{d x} (- 3 x))$

Step 2.4.3

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) e^{x^{3} - 3 x} (3 x^{2} - 3 \frac{d}{d x} x)$

Step 2.4.4

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) e^{x^{3} - 3 x} (3 x^{2} - 3 \cdot 1)$

Step 2.4.5

Multiply $- 3$ by $1$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) e^{x^{3} - 3 x} (3 x^{2} - 3)$

Step 2.5

Raise $3 x^{2} - 3$ to the power of $1$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) (3 x^{2} - 3) e^{x^{3} - 3 x}$

Step 2.6

Raise $3 x^{2} - 3$ to the power of $1$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x) + (3 x^{2} - 3) (3 x^{2} - 3) e^{x^{3} - 3 x}$

Step 2.7

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

$f'' (x) = e^{x^{3} - 3 x} (6 x) + {(3 x^{2} - 3)}^{1 + 1} e^{x^{3} - 3 x}$

Step 2.8

Add $1$ and $1$ .

$f'' (x) = e^{x^{3} - 3 x} (6 x) + {(3 x^{2} - 3)}^{2} e^{x^{3} - 3 x}$

Step 2.9

Reorder terms.

$f'' (x) = 6 e^{x^{3} - 3 x} x + e^{x^{3} - 3 x} {(3 x^{2} - 3)}^{2}$

Step 3

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

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

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x^{3} - 3 x$ .

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Step 4.1.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{3} - 3 x]$

Step 4.1.1.2

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

$e^{u} \frac{d}{d x} [x^{3} - 3 x]$

Step 4.1.1.3

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

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

Step 4.1.2

Differentiate.

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Step 4.1.2.1

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

$e^{x^{3} - 3 x} (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 3 x])$

Step 4.1.2.2

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

$e^{x^{3} - 3 x} (3 x^{2} + \frac{d}{d x} [- 3 x])$

Step 4.1.2.3

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

$e^{x^{3} - 3 x} (3 x^{2} - 3 \frac{d}{d x} [x])$

Step 4.1.2.4

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

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

Step 4.1.2.5

Multiply $- 3$ by $1$ .

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

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $e^{x^{3} - 3 x} (3 x^{2} - 3)$ .

$e^{x^{3} - 3 x} (3 x^{2} - 3)$

Step 5

Set the first derivative equal to $0$ then solve the equation $e^{x^{3} - 3 x} (3 x^{2} - 3) = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

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

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

$3 x^{2} - 3 = 0$

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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Step 5.3.2.1

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

$\ln (e^{x^{3} - 3 x}) = \ln (0)$

Step 5.3.2.2

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

Undefined

Step 5.3.2.3

There is no solution for $e^{x^{3} - 3 x} = 0$

No solution

Step 5.4

Set $3 x^{2} - 3$ equal to $0$ and solve for $x$ .

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Step 5.4.1

Set $3 x^{2} - 3$ equal to $0$ .

$3 x^{2} - 3 = 0$

Step 5.4.2

Solve $3 x^{2} - 3 = 0$ for $x$ .

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Step 5.4.2.1

Add $3$ to both sides of the equation.

$3 x^{2} = 3$

Step 5.4.2.2

Divide each term in $3 x^{2} = 3$ by $3$ and simplify.

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Step 5.4.2.2.1

Divide each term in $3 x^{2} = 3$ by $3$ .

$\frac{3 x^{2}}{3} = \frac{3}{3}$

Step 5.4.2.2.2

Simplify the left side.

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Step 5.4.2.2.2.1

Cancel the common factor of $3$ .

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Step 5.4.2.2.2.1.1

Cancel the common factor.

$\frac{3 x^{2}}{3} = \frac{3}{3}$

Step 5.4.2.2.2.1.2

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

$x^{2} = \frac{3}{3}$

Step 5.4.2.2.3

Simplify the right side.

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Step 5.4.2.2.3.1

Divide $3$ by $3$ .

$x^{2} = 1$

Step 5.4.2.3

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

$x = \pm \sqrt{1}$

Step 5.4.2.4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5.4.2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.4.2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Step 5.4.2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

Step 5.4.2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 5.5

The final solution is all the values that make $e^{x^{3} - 3 x} (3 x^{2} - 3) = 0$ true.

$x = 1, - 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 1, - 1$

Step 8

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

$6 e^{{(1)}^{3} - 3 \cdot 1} (1) + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Simplify each term.

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Step 9.1.1.1

One to any power is one.

$6 e^{1 - 3 \cdot 1} \cdot 1 + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.1.2

Multiply $- 3$ by $1$ .

$6 e^{1 - 3} \cdot 1 + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.2

Subtract $3$ from $1$ .

$6 e^{- 2} \cdot 1 + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.3

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

$6 \frac{1}{e^{2}} \cdot 1 + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.4

Combine $6$ and $\frac{1}{e^{2}}$ .

$\frac{6}{e^{2}} \cdot 1 + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.5

Multiply $\frac{6}{e^{2}}$ by $1$ .

$\frac{6}{e^{2}} + e^{{(1)}^{3} - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.6

Simplify each term.

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Step 9.1.6.1

One to any power is one.

$\frac{6}{e^{2}} + e^{1 - 3 \cdot 1} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.6.2

Multiply $- 3$ by $1$ .

$\frac{6}{e^{2}} + e^{1 - 3} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.7

Subtract $3$ from $1$ .

$\frac{6}{e^{2}} + e^{- 2} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.8

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

$\frac{6}{e^{2}} + \frac{1}{e^{2}} {(3 {(1)}^{2} - 3)}^{2}$

Step 9.1.9

Simplify each term.

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Step 9.1.9.1

One to any power is one.

$\frac{6}{e^{2}} + \frac{1}{e^{2}} {(3 \cdot 1 - 3)}^{2}$

Step 9.1.9.2

Multiply $3$ by $1$ .

$\frac{6}{e^{2}} + \frac{1}{e^{2}} {(3 - 3)}^{2}$

Step 9.1.10

Subtract $3$ from $3$ .

$\frac{6}{e^{2}} + \frac{1}{e^{2}} \cdot 0^{2}$

Step 9.1.11

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

$\frac{6}{e^{2}} + \frac{1}{e^{2}} \cdot 0$

Step 9.1.12

Multiply $\frac{1}{e^{2}}$ by $0$ .

$\frac{6}{e^{2}} + 0$

Step 9.2

Add $\frac{6}{e^{2}}$ and $0$ .

$\frac{6}{e^{2}}$

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.

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

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

One to any power is one.

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

Step 11.2.1.2

Multiply $- 3$ by $1$ .

$f (1) = e^{1 - 3}$

Step 11.2.2

Subtract $3$ from $1$ .

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

Step 11.2.3

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

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

Step 11.2.4

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

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

Step 12

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

$6 e^{{(- 1)}^{3} - 3 \cdot - 1} (- 1) + e^{{(- 1)}^{3} - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13

Evaluate the second derivative.

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Step 13.1

Simplify each term.

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Step 13.1.1

Simplify each term.

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Step 13.1.1.1

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

$6 e^{- 1 - 3 \cdot - 1} \cdot - 1 + e^{{(- 1)}^{3} - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.1.2

Multiply $- 3$ by $- 1$ .

$6 e^{- 1 + 3} \cdot - 1 + e^{{(- 1)}^{3} - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.2

Add $- 1$ and $3$ .

$6 e^{2} \cdot - 1 + e^{{(- 1)}^{3} - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.3

Multiply $- 1$ by $6$ .

$- 6 e^{2} + e^{{(- 1)}^{3} - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.4

Simplify each term.

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Step 13.1.4.1

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

$- 6 e^{2} + e^{- 1 - 3 \cdot - 1} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.4.2

Multiply $- 3$ by $- 1$ .

$- 6 e^{2} + e^{- 1 + 3} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.5

Add $- 1$ and $3$ .

$- 6 e^{2} + e^{2} {(3 {(- 1)}^{2} - 3)}^{2}$

Step 13.1.6

Simplify each term.

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Step 13.1.6.1

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

$- 6 e^{2} + e^{2} {(3 \cdot 1 - 3)}^{2}$

Step 13.1.6.2

Multiply $3$ by $1$ .

$- 6 e^{2} + e^{2} {(3 - 3)}^{2}$

Step 13.1.7

Subtract $3$ from $3$ .

$- 6 e^{2} + e^{2} \cdot 0^{2}$

Step 13.1.8

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

$- 6 e^{2} + e^{2} \cdot 0$

Step 13.1.9

Multiply $e^{2}$ by $0$ .

$- 6 e^{2} + 0$

Step 13.2

Add $- 6 e^{2}$ and $0$ .

$- 6 e^{2}$

Step 14

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

$x = - 1$ is a local maximum

Step 15

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

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Step 15.1

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

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

Step 15.2

Simplify the result.

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Step 15.2.1

Simplify each term.

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Step 15.2.1.1

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

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

Step 15.2.1.2

Multiply $- 3$ by $- 1$ .

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

Step 15.2.2

Add $- 1$ and $3$ .

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

Step 15.2.3

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

$y = e^{2}$

Step 16

These are the local extrema for $f (x) = e^{x^{3} - 3 x}$ .

$(1, \frac{1}{e^{2}})$ is a local minima

$(- 1, e^{2})$ is a local maxima

Step 17