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Question

(A) The function is defined as $f(x) = \begin{cases} x^2, & x  1 \end{cases}$.To check for differentiability

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$\{0, 1\}$

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(A) The function is defined as $f(x) = \begin{cases} x^2, & x  1 \end{cases}$.To check for differentiability at $x = 0$:Left-hand derivative $(LHD)$ at $x = 0$ is $\frac{d}{dx}(x^2) = 2x = 0$.Right-hand derivative $(RHD)$ at $x = 0$ is $\frac{d}{dx}(x) = 1$.Since $LHD 
eq RHD$,$f$ is not differentiable at $x = 0$.To check for differentiability at $x = 1$:Left-hand derivative $(LHD)$ at $x = 1$ is $\frac{d}{dx}(x) = 1$.Right-hand derivative $(RHD)$ at $x = 1$ is $\frac{d}{dx}(x^2) = 2x = 2$.Since $LHD 
eq RHD$,$f$ is not differentiable at $x = 1$.Thus,the set of points where $f$ is not differentiable is $S = \{0, 1\}$.

Let $f: R \rightarrow R$ be a function defined by $f(x) = \max \{x, x^2\}$. Let $S$ denote the set of all points in $R$ where $f$ is not differentiable. Then $S$ is:

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