Calculus Examples

$h (x) = e^{x^{2} - 4}$ on $- 2$ , $2$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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 $g (x) = x^{2} - 4$ .

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Step 1.1.1.1.1

To apply the Chain Rule, set $u$ as $x^{2} - 4$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{2} - 4]$

Step 1.1.1.1.2

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

$e^{u} \frac{d}{d x} [x^{2} - 4]$

Step 1.1.1.1.3

Replace all occurrences of $u$ with $x^{2} - 4$ .

$e^{x^{2} - 4} \frac{d}{d x} [x^{2} - 4]$

Step 1.1.1.2

Differentiate.

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Step 1.1.1.2.1

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

$e^{x^{2} - 4} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4])$

Step 1.1.1.2.2

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

$e^{x^{2} - 4} (2 x + \frac{d}{d x} [- 4])$

Step 1.1.1.2.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

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

Step 1.1.1.2.4

Add $2 x$ and $0$ .

$e^{x^{2} - 4} (2 x)$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Reorder the factors of $e^{x^{2} - 4} (2 x)$ .

$2 e^{x^{2} - 4} x$

Step 1.1.1.3.2

Reorder factors in $2 e^{x^{2} - 4} x$ .

$f' (x) = 2 x e^{x^{2} - 4}$

Step 1.1.2

The first derivative of $h (x)$ with respect to $x$ is $2 x e^{x^{2} - 4}$ .

$2 x e^{x^{2} - 4}$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $2 x e^{x^{2} - 4} = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$2 x e^{x^{2} - 4} = 0$

Step 1.2.2

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

$x = 0$

$e^{x^{2} - 4} = 0$

Step 1.2.3

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.4

Set $e^{x^{2} - 4}$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $e^{x^{2} - 4}$ equal to $0$ .

$e^{x^{2} - 4} = 0$

Step 1.2.4.2

Solve $e^{x^{2} - 4} = 0$ for $x$ .

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Step 1.2.4.2.1

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

$\ln (e^{x^{2} - 4}) = \ln (0)$

Step 1.2.4.2.2

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

Undefined

Step 1.2.4.2.3

There is no solution for $e^{x^{2} - 4} = 0$

No solution

Step 1.2.5

The final solution is all the values that make $2 x e^{x^{2} - 4} = 0$ true.

$x = 0$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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 1.4

Evaluate $e^{x^{2} - 4}$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 0$ .

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Step 1.4.1.1

Substitute $0$ for $x$ .

$e^{{(0)}^{2} - 4}$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

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

$e^{0 - 4}$

Step 1.4.1.2.2

Subtract $4$ from $0$ .

$e^{- 4}$

Step 1.4.1.2.3

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

$\frac{1}{e^{4}}$

Step 1.4.2

List all of the points.

$(0, \frac{1}{e^{4}})$

Step 2

Evaluate at the included endpoints.

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Step 2.1

Evaluate at $x = - 2$ .

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Step 2.1.1

Substitute $- 2$ for $x$ .

$e^{{(- 2)}^{2} - 4}$

Step 2.1.2

Simplify.

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Step 2.1.2.1

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

$e^{4 - 4}$

Step 2.1.2.2

Subtract $4$ from $4$ .

$e^{0}$

Step 2.1.2.3

Anything raised to $0$ is $1$ .

$1$

Step 2.2

Evaluate at $x = 2$ .

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Step 2.2.1

Substitute $2$ for $x$ .

$e^{{(2)}^{2} - 4}$

Step 2.2.2

Simplify.

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Step 2.2.2.1

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

$e^{4 - 4}$

Step 2.2.2.2

Subtract $4$ from $4$ .

$e^{0}$

Step 2.2.2.3

Anything raised to $0$ is $1$ .

$1$

Step 2.3

List all of the points.

$(- 2, 1), (2, 1)$

Step 3

Compare the $h (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $h (x)$ value and the minimum will occur at the lowest $h (x)$ value.

Absolute Maximum: $(- 2, 1), (2, 1)$

Absolute Minimum: $(0, \frac{1}{e^{4}})$

Step 4