Q. 80 - 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.

$f\left(x\right)=\frac{1}{x+3}$

Step-by-Step Solution

Answer

The value of $\frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{(x+3)(x+h+3)}$

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
$f\left(x\right)=\frac{1}{x+3}$

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.

$f(x+h)=\frac{1}{x+h+3}$

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

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{1}{x+h+3}-\frac{1}{x+3}\right) \end{gathered}$$

The LCD of $x+h+3 \mathrm{and} a+3 \mathrm{is} (x+3)(x+h+3)$ so multiply with the LCD.

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{1}{x+h+3}·\frac{(x+3)(x+h+3)}{(x+3)(x+h+3)}-\frac{1}{x+3}·\frac{(x+3)(x+h+3)}{(x+3)(x+h+3)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x+3}{(x+3)(x+h+3)}-\frac{x+h+3}{(x+3)(x+h+3)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x+3-\left(x+h+3\right)}{(x+3)(x+h+3)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x+3-x-h-3}{(x+3)(x+h+3)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}·\frac{h}{(x+3)(x+h+3)} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{(x+3)(x+h+3)} \end{gathered}$$