Question

Let $f:(1,\infty )\to (1,\infty )$ is defined as $f\left(x\right)=\frac{x^2}{x^2-x}$. The inverse function of $f\left(x\right)$ is

• A. $\frac{1}{x-1}$
• B. $\frac{x}{x-1}$
• C. $\frac{1}{1-x}$
• D. $\frac{x}{1-x}$

Answer 1

The correct option is D $\frac{x}{1-x}$

$f:(1,\infty )\to (1,\infty )f\left(x\right)=\frac{x^2}{x^2-x}y=\frac{x^2}{x^2-x}=\frac{x^2}{x^2(1-\frac{1}{x^2})}y=\frac{x}{x-1}xy-y=xy=x(y-1)Replacey\to xx=\frac{y}{y-1}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{x}{x-1}$