Question

Let the functions \(f:\mathbb R\to\mathbb R\) and \(g:\mathbb R\to\mathbb R\) be defined as :
\(f(x)=\begin{cases}x+2, & x<0\\ x^2, & x\ge0\end{cases}\) and
\(g(x)=\begin{cases}x^3, & x<1\\ 3x-2, & x\ge1\end{cases}\)
Then, the number of points in \(\mathbb R\) where \((f\circ g)(x)\) is NOT differentiable is equal to :

• A. 0
• B. 3
• C. 1
• D. 2

Hint:

Find \(f\circ g\) piecewise: it is \(x^3+2\) for \(x<0\), \(x^6\) for \(0\le x<1\), \((3x-2)^2\) for \(x\ge1\); check the junctions 0 and 1.

Answer 1

The Correct Option is C

For \(x<1\), \(g(x)=x^3\): if \(x<0\), \(g<0\) so \(f(g)=x^3+2\); if \(0\le x<1\), \(g\ge0\) so \(f(g)=x^6\).
For \(x\ge1\), \(g=3x-2\ge1>0\), so \(f(g)=(3x-2)^2\).
\((f\circ g)(x)=x^3+2\ (x<0),\ x^6\ (0\le x<1),\ (3x-2)^2\ (x\ge1)\).
At \(x=0\): left limit \(=2\), value \(=0\): discontinuous, hence NOT differentiable.
At \(x=1\): left limit \(=1\), right limit \(=1\), value \(=1\); left derivative \(6x^5=6\), right derivative \(6(3x-2)=6\): differentiable.
Each piece is differentiable elsewhere. Only 1 point.
Answer: (3) 1