Question

Find the value of $f\left(a\right),f\left(a+h\right),\frac{f\left(a+h\right)-f\left(a\right)}{h}$for the function $f\left(x\right)=\frac{1}{x+1}$.

Short answer

The value of $f\left(a\right)=\frac{1}{a+1},f\left(a+h\right)=\frac{1}{a+h+1},\frac{f\left(a+h\right)-f\left(a\right)}{h}=-\frac{1}{\left(a+1\right)\left(a+h+1\right)}$.

Step-by-step solution

Step 1. Evaluate the values of fa and fa+h.

From the given function,

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

Evaluate $f\left(a\right)\text{and}f\left(a+h\right)$gives,

role="math" localid="1644565860619" $$\begin{gathered} f\left(a\right)=\frac{1}{\left(a\right)+1}\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =\frac{1}{a+1} \\ f\left(a+h\right)=\frac{1}{\left(a+h\right)+1}\mathrm{....}\left(\text{Substitute}x=a+h\text{in}\left(1\right)\right) \\ =\frac{1}{a+h+1} \end{gathered}$$

Step 2. Subtract fa from fa+h and divide by h where h≠0.

Evaluate $\frac{f\left(a+h\right)-f\left(a\right)}{h}$gives,

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

Step 3. Description of steps.

The value of $f\left(a\right)=\frac{1}{a+1},f\left(a+h\right)=\frac{1}{a+h+1},\frac{f\left(a+h\right)-f\left(a\right)}{h}=-\frac{1}{\left(a+1\right)\left(a+h+1\right)}$.