Number of points where the function f(x) = Maximum {sgn (x), $-\sqrt{9-x^2}, x^3$} is continuous but not differentiable, is
Answer & explanation
Correct answer: option 3
We observe that $g(x)=-\sqrt{9-x^2}$ is defined for $x \in[-3,3]$
So, f(x) is defined on [-3, 3]
In the graph, continuous curve shows the graph of f(x).
Clearly, we have
$f(x)= \begin{cases}-\sqrt{9-x^2}& , \quad -3 \leq x \leq-2 \sqrt{2} \\ -1 & , \quad-2 \sqrt{2} \leq x \leq-1 \\ x^3 & , \quad-1 \leq x \leq 0 \\ 1 & , \quad 0<x \leq 1 \\ x^3 & , \quad 3 \geq x \geq 1\end{cases}$
It is evident from the graph that f(x) is continuous but not differentiable at A, B, C and Q.
Thus, there are 4 points where f(x) is continuous but not differentiable. It is to note that at point P, right hand derivate $-\infty$ and at Q the left hand derivative is $+\infty$.
So, f(x) is not differentiable at P and Q.