Question

If $f\left(x\right)=\frac{x-1}{x+1}$, then show that

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

Answer 1

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

(i) Replacing x by $\frac{1}{x}$ in (1), we get
$$\begin{gathered} f\left(\frac{1}{x}\right)=\frac{{\displaystyle \frac{1}{x}}-1}{{\displaystyle \frac{1}{x}}+1} \\ =\frac{1-x}{1+x} \\ =-\frac{x-1}{x+1} \\ =-f\left(x\right) \end{gathered}$$

(ii) Replacing x by $-\frac{1}{x}$ in (1), we get
$$\begin{gathered} f\left(-\frac{1}{x}\right)=\frac{-{\displaystyle \frac{1}{x}}-1}{-{\displaystyle \frac{1}{x}}+1} \\ =\frac{-1-x}{-1+x} \\ =-\frac{x+1}{x-1} \\ =-\frac{1}{\left({\displaystyle \frac{x-1}{x+1}}\right)} \\ =-\frac{1}{f\left(x\right)} \end{gathered}$$