Precalculus Examples

$f (x) = 8 x - 3$ , $x = 1$

Consider the difference quotient formula. Use this formula to solve for the slope of the secant line $m_{s e c}$ .

$m_{s e c} = \frac{f (x + h) - f (x)}{h}$

Find the components of the definition.

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

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

$f (x + h) = 8 (x + h) - 3$

Simplify the result.

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Apply the distributive property.

$f (x + h) = 8 x + 8 h - 3$

The final answer is $8 x + 8 h - 3$ .

$8 x + 8 h - 3$

Find the components of the definition.

$f (x + h) = 8 x + 8 h - 3$

$f (x) = 8 x - 3$

$f (x + h) = 8 x + 8 h - 3$

$f (x) = 8 x - 3$

Plug in the components.

$\frac{f (x + h) - f (x)}{h} = \frac{8 x + 8 h - 3 - (8 x - 3)}{h}$

Simplify.

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Simplify the numerator.

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Apply the distributive property.

$\frac{8 x + 8 h - 3 + (- (8 x) - - 3)}{h}$

Multiply $8$ by $- 1$ to get $- 8$ .

$\frac{8 x + 8 h - 3 + (- 8 x - - 3)}{h}$

Multiply $- 1$ by $- 3$ to get $3$ .

$\frac{8 x + 8 h - 3 + (- 8 x + 3)}{h}$

Remove parentheses.

$\frac{8 x + 8 h - 3 - 8 x + 3}{h}$

Subtract $8 x$ from $8 x$ to get $0$ .

$\frac{8 h - 3 + 0 + 3}{h}$

Add $8 h$ and $0$ to get $8 h$ .

$\frac{8 h - 3 + 3}{h}$

Add $- 3$ and $3$ to get $0$ .

$\frac{8 h + 0}{h}$

Add $8 h$ and $0$ to get $8 h$ .

$\frac{8 h}{h}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{8 h}{h}$

Divide $8$ by $1$ to get $8$ .

$8$

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