GATE CS 2024 Set 1 Q1 — Calculus: Limits, Continuity, Differentiability

To find the points where the function $f(x) = \max\{x, x^3\}$ is not differentiable, we first determine which function is larger in different intervals of the real line.

We find the intersection points of $y = x$ and $y = x^3$ by solving:

These points partition the real line into four intervals. Let's evaluate which function is larger in each interval:

• For $x < -1$: Since $x > x^3$ (e.g., $-2 > -8$), we have $f(x) = x$.
• For $-1 \le x < 0$: Since $x^3 > x$ (e.g., $-0.125 > -0.5$), we have $f(x) = x^3$.
• For $0 \le x < 1$: Since $x > x^3$ (e.g., $0.5 > 0.125$), we have $f(x) = x$.
• For $x \ge 1$: Since $x^3 > x$ (e.g., $8 > 2$), we have $f(x) = x^3$.

Thus, we can write $f(x)$ as a piecewise function:

Now, we check the differentiability at the transition points $x = -1, 0, 1$ by comparing the Left-Hand Derivative (LHD) and Right-Hand Derivative (RHD):

1. At $x = -1$:
   $\text{LHD} = \frac{d}{dx}(x)\Big|_{x=-1} = 1$
   $\text{RHD} = \frac{d}{dx}(x^3)\Big|_{x=-1} = 3(-1)^2 = 3$
   Since $\text{LHD} \neq \text{RHD}$, $f(x)$ is not differentiable at $x = -1$.


2. At $x = 0$:
   $\text{LHD} = \frac{d}{dx}(x^3)\Big|_{x=0} = 3(0)^2 = 0$
   $\text{RHD} = \frac{d}{dx}(x)\Big|_{x=0} = 1$
   Since $\text{LHD} \neq \text{RHD}$, $f(x)$ is not differentiable at $x = 0$.


3. At $x = 1$:
   $\text{LHD} = \frac{d}{dx}(x)\Big|_{x=1} = 1$
   $\text{RHD} = \frac{d}{dx}(x^3)\Big|_{x=1} = 3(1)^2 = 3$
   Since $\text{LHD} \neq \text{RHD}$, $f(x)$ is not differentiable at $x = 1$.

Therefore, the set of all points where $f(x)$ is not differentiable is $\{-1, 0, 1\}$.