Chapter 5.7, Problem 29WE

To determine

To express $\frac{f(x+h)-f(x)}{h}$ as a single simplified fraction.

Answer to Problem 29WE

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

Explanation of Solution

Given:

Function: $f(x)=\frac{1}{x+1}$

Concept used:

To find $\frac{f(x+h)-f(x)}{h}$ , the following steps are to be followed:

1. Find $f(x+h)$ by substituting $(x+h)$ in the place of x in the given function.
2. Substitute in $f(x+h)-f(x)$
3. Divide by h .

Calculation:

Given function is $f(x)=\frac{1}{x+1}.$

Find $f(x+h)$ by substituting $(x+h)$ in the place of x in $f(x)$ .

  $f(x+h)=\frac{1}{(x+h)+1}$

Substitute in $\frac{f(x+h)-f(x)}{h},$

  $\begin{array}{l}\Rightarrow \frac{\frac{1}{(x+h)+1}-\frac{1}{(x+h)}}{h}\\ \Rightarrow \frac{\frac{1\times (x+h)}{(x+h+1)\times (x+h)}-\frac{1\times (x+h+1)}{(x+h)\times (x+h+1)}}{h}\\ \Rightarrow \frac{\frac{x+h}{(x+h)(x+h+1)}-\frac{x+h+1}{(x+h)(x+h+1)}}{h}\\ \Rightarrow \frac{\frac{(x+h)-(x+h+1)}{(x+h)(x+h+1)}}{h}\\ \Rightarrow \frac{\frac{-1}{(x+h)(x+h+1)}}{h}\\ \Rightarrow \frac{-1}{h(x+h)(x+h+1)}\end{array}$

Conclusion:

Therefore, the value of $\frac{f(x+h)-f(x)}{h}=\frac{-1}{h(x+h)(x+h+1)}.$