Question

Let a function, \(f:\mathbb{R}\to\mathbb{R}\) be defined by \(f(x)=1+x|x|\) and let \(\operatorname{sgn}(x)=\begin{cases}1, & \text{if } x\ge 0\\ -1, & \text{if } x<0\end{cases}\)
Then :

• A. \(f\) is not invertible
• B. \(f^{-1}(x)=\operatorname{sgn}(x-1)\sqrt{|x-1|}\)
• C. \(f^{-1}(x)=\sqrt{|x-1|}\)
• D. \(f^{-1}(x)=\operatorname{sgn}(x)\sqrt{|x-1|}\)

Hint:

Write \(f\) piecewise (\(1+x^2\) for \(x\ge0\), \(1-x^2\) for \(x<0\)), show it is strictly increasing and onto, then solve \(y=1+x|x|\) for \(x\).

Answer 1

The Correct Option is B

For \(x\ge 0\), \(f(x)=1+x^2\ge 1\); for \(x<0\), \(f(x)=1-x^2<1\).
Both pieces are strictly increasing; the left piece takes all values below 1 and the right piece takes all values from 1 upwards, so \(f\) is one-one and onto \(\mathbb{R}\). Hence \(f\) is invertible and option (1) is false.
Let \(y=1+x|x|\). Then \(y-1=x|x|\).
If \(y\ge 1\), then \(x\ge 0\) and \(y-1=x^2\), so \(x=\sqrt{y-1}\).
If \(y<1\), then \(x<0\) and \(y-1=-x^2\), so \(x=-\sqrt{1-y}\).
Both cases are written together as \(x=\operatorname{sgn}(y-1)\sqrt{|y-1|}\).
Therefore \(f^{-1}(x)=\operatorname{sgn}(x-1)\sqrt{|x-1|}\).
Answer: (2) \(f^{-1}(x)=\operatorname{sgn}(x-1)\sqrt{|x-1|}\)