Question

If $f\left(x\right)=\frac{x-1}{x+1}$ then find the value of $f\left(2x\right)$.

Answer 1

Step 1: Solve for the value of $x$.

Given: $f\left(x\right)=\frac{x-1}{x+1}$

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

$xf\left(x\right)+f\left(x\right)=x-1$

$1+f\left(x\right)=x-xf\left(x\right)$

$1+f\left(x\right)=x\left[1-f\left(x\right)\right]$

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

Step 2: Compute the required value.

Substituting $2x$ in place of $x$

$$\begin{gathered} f\left(2x\right)=\frac{2x-1}{2x+1} \\ f\left(2x\right)=\frac{2\left[{\displaystyle \frac{1+f\left(x\right)}{1-f\left(x\right)}}\right]-1}{2\left[{\displaystyle \frac{1+f\left(x\right)}{1-f\left(x\right)}}\right]+1} \\ f\left(2x\right)=\frac{2+2f\left(x\right)+f\left(x\right)-1}{2+2f\left(x\right)-f\left(x\right)+1} \\ f\left(2x\right)=\frac{1+3f\left(x\right)}{3+f\left(x\right)} \end{gathered}$$

Hence, the value of $f\left(2x\right)$ is $\frac{1+3f\left(x\right)}{3+f\left(x\right)}$.