Question

The function of $f\left(x\right)=\frac{1}{\sqrt{x+1}}$ is defined for $\{x|x$ is a real number and $x>-1\}$. Write an expression for the inverse of $f\left(x\right)$.

• A. $\frac{1-x^2}{x^2}$ such that $x>0$
• B. $\frac{1-x}{x}$
• C. $\frac{x^2-1}{x^2}$
• D. $\frac{1-x^2}{x^2}$ such that $x>-1$
• E. none of these

Answer 1

The correct option is A $\frac{1-x^2}{x^2}$ such that $x>0$

Given, $f\left(x\right)=\frac{1}{\sqrt{x+1}}$

Let it be $y$, we have $f^{-1}\left(y\right)=x$
We get $y=\frac{1}{\sqrt{x+1}}$

$\Rightarrow \sqrt{x+1}=\frac{1}{y}$
$\Rightarrow x=\frac{1}{y^2}-1=\frac{1-y^2}{y^2}=f^{-1}\left(y\right)$
Therefore, we get $f^{-1}\left(x\right)=\frac{1-y^2}{y^2}$, such that $x>0$