Q. 82 - 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.

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

Step-by-Step Solution

Answer

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

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

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.

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

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

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

To rationalize we multiply and divide with $\sqrt{x+h+1}+\sqrt{x+1}$

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

Using the formula $(a^2-b^2)=(a+b)(a-b)$

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