Q. 77 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for each function. Be sure to simplify.

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

In this problems we have to find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for given function. Be sure to simplify.

The given function is
$f\left(x\right)=x^2-x+4$

Step 2. Firstly we find the value of f ( x + h ) We find the value of f ( x + h ) by putting x = x + h in given function.

$f(x+h)={(x+h)}^2-(x+h)+4$

Using the formula ${(a+b)}^2=a^2+2ab+b^2$

$f(x+h)={x^2+2xh+h}^2-x-h+4$

Step 3. Putting the value of f ( x )   and   f ( x + h ) in the f ( x + h ) -   f ( x ) h

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{{x^2+2xh+h}^2-x-h+4-(x^2-x+4)}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{{x^2+2xh+h}^2-x-h+4-x^2+x-4}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{{2xh+h}^2-h}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{h(2x+h-1)}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=2x+h-1 \end{gathered}$$