GATE CS 2024 Set 1 — Question 11

The function is given by $f(x) = \max\{x, x^3\}$.
To analyze the function, we find the points where the graphs of $y=x$ and $y=x^3$ intersect:$$x = x^3 \implies x^3 - x = 0 \implies x(x-1)(x+1) = 0$$The intersection points are $x = -1, 0, 1$.We determine the value of $f(x)$ in the intervals defined by these points:Thus, the function can be written piecewise as:$$f(x) = \begin{cases} x & x \le -1 \\ x^3 & -1 < x \le 0 \\ x & 0 < x \le 1 \\ x^3 & x > 1 \end{cases}$$We check for differentiability at the transition points $x = -1, 0, 1$ by comparing the Left Hand Derivative (LHD) and Right Hand Derivative (RHD).At $x = -1$:
LHD $= \frac{d}{dx}(x) = 1$
RHD $= \frac{d}{dx}(x^3)\big|_{x=-1} = 3(-1)^2 = 3$
Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = -1$.At $x = 0$:
LHD $= \frac{d}{dx}(x^3)\big|_{x=0} = 3(0)^2 = 0$
RHD $= \frac{d}{dx}(x) = 1$
Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 0$.At $x = 1$:
LHD $= \frac{d}{dx}(x) = 1$
RHD $= \frac{d}{dx}(x^3)\big|_{x=1} = 3(1)^2 = 3$
Since LHD $\neq$ RHD, $f(x)$ is not differentiable at $x = 1$.The set of points where $f(x)$ is not differentiable is $\{-1, 0, 1\}$.