Chapter 2.1, Problem 77AYU

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.

$f\left(x\right)\text{ = }{x}^2\text{ - }x\text{ + 4}$

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 $f\left(x\right)=x²-x+4$

Answer to Problem 77AYU

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\text{ 2}x+h-\text{ 1}$

Explanation of Solution

Given:

The function $f\left(x\right)=x²-x+4$

Calculation:

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{{(x+h)}^2-\left(x+h\right)+4-\left(x^2-x+4\right)}{h}$

$=\frac{\left(x^2+2xh+h^2\right)-\left(x+h\right)+4-x^2+x-4}{h}$

$=\frac{x^2+2xh+h^2-x-h+4-x^2+x-4}{h}$

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

$=\frac{h\left(2x+h-1\right)}{h}$

$=\text{ 2}x+h-\text{ 1}$