Question

If $f\left(x\right)=\frac{x-1}{x+1},$ then $f(2x)$ is equal to

• A. $\frac{f\left(x\right)+1}{f\left(x\right)+3}$
• B. $\frac{3f\left(x\right)+1}{f\left(x\right)+3}$
• C. $\frac{f\left(x\right)+3}{f\left(x\right)+1}$
• D. $\frac{f\left(x\right)+3}{3f\left(x\right)+1}$

Answer 1

Answer: (B)

Given $f\left(x\right)=\frac{x-1}{x+1}$
$\therefore f\left(2x\right)=\frac{2x-1}{2x+1}$
$=\frac{2\left(2x-1\right)}{2\left(2x+1\right)}$ (multiply and divide by 2)
$=\frac{4x-2}{4x+2}=\frac{3x+x-3+1}{3x+x+3-1}=\frac{3\left(x-1\right)+x+1}{3\left(x+1\right)+x-1}$
$=\frac{3\left[\frac{x-1}{x+1}\right]+1}{\frac{x-1}{x+1}+3}=\frac{3f\left(x\right)+1}{f\left(x\right)+3}$