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 role="math" localid="1644567776550" $f\left(x\right)=\frac{2x}{x-1}$.

Short answer

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

Step-by-step solution

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

From the given function,

$f\left(x\right)=\frac{2x}{x-1}\mathrm{....}\left(1\right)$

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

role="math" localid="1644568144405" $$\begin{gathered} f\left(a\right)=\frac{2\left(a\right)}{\left(a\right)-1}\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =\frac{2a}{a-1} \\ f\left(a+h\right)=\frac{2\left(a+h\right)}{\left(a+h\right)-1}\mathrm{....}\left(\text{Substitute}x=a+h\text{in}\left(1\right)\right) \\ =\frac{2a+2h}{a+h-1} \end{gathered}$$

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,

role="math" localid="1644568537135" $$\begin{gathered} \frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{\frac{2a+2h}{a+h-1}-\frac{2a}{a-1}}{h} \\ =\frac{2a^2-2a+2ah-2h-2a^2-2ah+2a}{h\left(a-1\right)\left(a+h-1\right)} \\ =\frac{-2}{\left(a-1\right)\left(a+h-1\right)} \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

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