Chapter 2.1, Problem 78AYU

In Problems 79-90, find the difference quotient of $f$; that is, find $\frac{f\left(x\text{ + }h\right)\text{ - }f\left(x\right)}{h}$, $h\ne \text{ 0}$, for each function. Be sure to simplify.

$f\left(x\right)\text{ = 3}{x}^2\text{ - 2}x\text{ + 6}$

To determine

To find: The difference function of $f$, that is $\frac{f\left(x+h\right)-f\left(x\right)}{h}$, $h\ne 0$ for the function $f\left(x\right)=\text{ 3}x²-\text{ 2}x+\text{ 6}$

Answer to Problem 78AYU

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\text{ 6}x+\text{ 3}h-\text{ 2}$

Explanation of Solution

Given:

The function $f\left(x\right)=\text{ 3}x²-\text{ 2}x+\text{ 6}$

Calculation:

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{3{(x+h)}^2-2\left(x+h\right)+6-\left(3x^2-2x+6\right)}{h}$

$=\frac{3\left(x^2+2xh+h^2\right)-2\left(x+h\right)+6-3x^2+2x-6}{h}$

$=\frac{3x^2+6xh+3h^2-2x-2h+6-3x^2+2x-6}{h}$

$=\frac{6xh+3h^2-2h}{h}$

$=\frac{h\left(6x+3h-2\right)}{h}$

$=\text{ 6}x+\text{ 3}h-\text{ 2}$