Question

If $f:R->R$ is defined by $f\left(x\right)=|x|$, then

• A. $f^{-1}\left(x\right)=-x$
• B. $f^{-1}\left(x\right)=\frac{1}{|x|}$
• C. The function $f^{-1}\left(x\right)$ does not exist
• D. $f^{-1}\left(x\right)=\frac{1}{x}$

Answer 1

The correct option is C The function $f^{-1}\left(x\right)$ does not exist

For a function $f\left(x\right)$ to be invertible, the function must be one-one and onto.

The range of $f\left(x\right)=|x|$ is $[0,\infty )$, while the co-domain of $f\left(x\right)$ is given as $R$. Hence $f\left(x\right)$ is not onto.

Also, since $f\left(x\right)=f(-x)$, $f\left(x\right)$ is also not one-one in its domain.

Hence, f(x) is not invertible, ie, the function $f^{-1}\left(x\right)$ does not exist.

Option C is the right answer.