Precalculus Examples

$f (x) = 18 + 3 e^{- 3 x}$

Step 1

Consider the difference quotient formula.

$\frac{f (x + h) - f (x)}{h}$

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at $x = x + h$ .

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Step 2.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = 18 + 3 e^{- 3 (x + h)}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Apply the distributive property.

$f (x + h) = 18 + 3 e^{- 3 x - 3 h}$

Step 2.1.2.2

The final answer is $18 + 3 e^{- 3 x - 3 h}$ .

$18 + 3 e^{- 3 x - 3 h}$

Step 2.2

Find the components of the definition.

$f (x + h) = 18 + 3 e^{- 3 x - 3 h}$

$f (x) = 18 + 3 e^{- 3 x}$

$f (x + h) = 18 + 3 e^{- 3 x - 3 h}$

$f (x) = 18 + 3 e^{- 3 x}$

Step 3

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{18 + 3 e^{- 3 x - 3 h} - (18 + 3 e^{- 3 x})}{h}$

Step 4

Simplify the numerator.

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Step 4.1

Apply the distributive property.

$\frac{18 + 3 e^{- 3 x - 3 h} - 1 \cdot 18 - (3 e^{- 3 x})}{h}$

Step 4.2

Multiply $- 1$ by $18$ .

$\frac{18 + 3 e^{- 3 x - 3 h} - 18 - (3 e^{- 3 x})}{h}$

Step 4.3

Multiply $3$ by $- 1$ .

$\frac{18 + 3 e^{- 3 x - 3 h} - 18 - 3 e^{- 3 x}}{h}$

Step 4.4

Subtract $18$ from $18$ .

$\frac{3 e^{- 3 x - 3 h} + 0 - 3 e^{- 3 x}}{h}$

Step 4.5

Add $3 e^{- 3 x - 3 h}$ and $0$ .

$\frac{3 e^{- 3 x - 3 h} - 3 e^{- 3 x}}{h}$

Step 4.6

Rewrite $3 e^{- 3 x - 3 h} - 3 e^{- 3 x}$ in a factored form.

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Step 4.6.1

Factor $3$ out of $3 e^{- 3 x - 3 h} - 3 e^{- 3 x}$ .

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Step 4.6.1.1

Factor $3$ out of $3 e^{- 3 x - 3 h}$ .

$\frac{3 e^{- 3 x - 3 h} - 3 e^{- 3 x}}{h}$

Step 4.6.1.2

Factor $3$ out of $- 3 e^{- 3 x}$ .

$\frac{3 e^{- 3 x - 3 h} + 3 (- e^{- 3 x})}{h}$

Step 4.6.1.3

Factor $3$ out of $3 e^{- 3 x - 3 h} + 3 (- e^{- 3 x})$ .

$\frac{3 (e^{- 3 x - 3 h} - e^{- 3 x})}{h}$

Step 4.6.2

Rewrite $e^{- 3 x - 3 h}$ as ${(e^{- x - h})}^{3}$ .

$\frac{3 ({(e^{- x - h})}^{3} - e^{- 3 x})}{h}$

Step 4.6.3

Rewrite $e^{- 3 x}$ as ${(e^{- x})}^{3}$ .

$\frac{3 ({(e^{- x - h})}^{3} - {(e^{- x})}^{3})}{h}$

Step 4.6.4

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = e^{- x - h}$ and $b = e^{- x}$ .

$\frac{3 ((e^{- x - h} - e^{- x}) ({(e^{- x - h})}^{2} + e^{- x - h} e^{- x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5

Simplify.

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Step 4.6.5.1

Multiply the exponents in ${(e^{- x - h})}^{2}$ .

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Step 4.6.5.1.1

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

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{(- x - h) \cdot 2} + e^{- x - h} e^{- x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.1.2

Apply the distributive property.

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- x \cdot 2 - h \cdot 2} + e^{- x - h} e^{- x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.1.3

Multiply $2$ by $- 1$ .

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - h \cdot 2} + e^{- x - h} e^{- x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.1.4

Multiply $2$ by $- 1$ .

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- x - h} e^{- x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.2

Multiply $e^{- x - h}$ by $e^{- x}$ by adding the exponents.

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Step 4.6.5.2.1

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

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- x - h - x} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.2.2

Subtract $x$ from $- x$ .

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- 2 x - h} + {(e^{- x})}^{2}))}{h}$

Step 4.6.5.3

Multiply the exponents in ${(e^{- x})}^{2}$ .

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Step 4.6.5.3.1

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

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- 2 x - h} + e^{- x \cdot 2}))}{h}$

Step 4.6.5.3.2

Multiply $2$ by $- 1$ .

$\frac{3 ((e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- 2 x - h} + e^{- 2 x}))}{h}$

$\frac{3 (e^{- x - h} - e^{- x}) (e^{- 2 x - 2 h} + e^{- 2 x - h} + e^{- 2 x})}{h}$

Step 5