Precalculus Examples

$f (x) = - 2 x + 12 x - 9$

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.

$f (x + h) = - 2 (x + h) + 12 (x + h) - 9$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Simplify each term.

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Step 2.1.2.1.1

Apply the distributive property.

$f (x + h) = - 2 x - 2 h + 12 (x + h) - 9$

Step 2.1.2.1.2

Apply the distributive property.

$f (x + h) = - 2 x - 2 h + 12 x + 12 h - 9$

$f (x + h) = - 2 x - 2 h + 12 x + 12 h - 9$

Step 2.1.2.2

Simplify by adding terms.

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Step 2.1.2.2.1

Add $- 2 x$ and $12 x$ .

$f (x + h) = - 2 h + 10 x + 12 h - 9$

Step 2.1.2.2.2

Add $- 2 h$ and $12 h$ .

$f (x + h) = 10 h + 10 x - 9$

$f (x + h) = 10 h + 10 x - 9$

Step 2.1.2.3

The final answer is $10 h + 10 x - 9$ .

$10 h + 10 x - 9$

$10 h + 10 x - 9$

$10 h + 10 x - 9$

Step 2.2

Find the components of the definition.

$f (x + h) = 10 h + 10 x - 9$

$f (x) = 10 x - 9$

$f (x + h) = 10 h + 10 x - 9$

$f (x) = 10 x - 9$

Step 3

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{10 h + 10 x - 9 - (10 x - 9)}{h}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Apply the distributive property.

$\frac{10 h + 10 x - 9 - (10 x) - - 9}{h}$

Step 4.1.2

Multiply $10$ by $- 1$ .

$\frac{10 h + 10 x - 9 - 10 x - - 9}{h}$

Step 4.1.3

Multiply $- 1$ by $- 9$ .

$\frac{10 h + 10 x - 9 - 10 x + 9}{h}$

Step 4.1.4

Subtract $10 x$ from $10 x$ .

$\frac{10 h + 0 - 9 + 9}{h}$

Step 4.1.5

Add $10 h$ and $0$ .

$\frac{10 h - 9 + 9}{h}$

Step 4.1.6

Add $- 9$ and $9$ .

$\frac{10 h + 0}{h}$

Step 4.1.7

Add $10 h$ and $0$ .

$\frac{10 h}{h}$

$\frac{10 h}{h}$

Step 4.2

Cancel the common factor of $h$ .

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Step 4.2.1

Cancel the common factor.

$\frac{10 h}{h}$

Step 4.2.2

Divide $10$ by $1$ .

$10$

$10$

$10$

Step 5

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$