Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) = \max\{x, x^3\}$, $x \in \mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers. The set of all points where $f(x)$ is NOT differentiable is

To determine where $f(x) = \max\{x, x^3\}$ is not differentiable, we analyze the sign of $x^3 - x = x(x-1)(x+1)$ across different intervals:

1. For $x \in (-\infty, -1)$: $x^3 < x \implies f(x) = x$

2. For $x \in (-1, 0)$: $x^3 > x \implies f(x) = x^3$

3. For $x \in (0, 1)$: $x > x^3 \implies f(x) = x$

4. For $x \in (1, \infty)$: $x^3 > x \implies f(x) = x^3$

Now, check differentiability at the transition points $x = -1, 0, 1$:

• At $x = -1$:

  • Left-hand derivative: $f^{\prime}(-1^-) = \left.\frac{d}{dx}(x)\right|_{x=-1} = 1$

  • Right-hand derivative: $f^{\prime}(-1^+) = \left.\frac{d}{dx}(x^3)\right|_{x=-1} = 3(-1)^2 = 3$

  Since $f^{\prime}(-1^-) \neq f^{\prime}(-1^+)$, $f$ is not differentiable at $x = -1$.

• At $x = 0$:

  • Left-hand derivative: $f^{\prime}(0^-) = \left.\frac{d}{dx}(x^3)\right|_{x=0} = 3(0)^2 = 0$

  • Right-hand derivative: $f^{\prime}(0^+) = \left.\frac{d}{dx}(x)\right|_{x=0} = 1$

  Since $f^{\prime}(0^-) \neq f^{\prime}(0^+)$, $f$ is not differentiable at $x = 0$.

• At $x = 1$:

  • Left-hand derivative: $f^{\prime}(1^-) = \left.\frac{d}{dx}(x)\right|_{x=1} = 1$

  • Right-hand derivative: $f^{\prime}(1^+) = \left.\frac{d}{dx}(x^3)\right|_{x=1} = 3(1)^2 = 3$

  Since $f^{\prime}(1^-) \neq f^{\prime}(1^+)$, $f$ is not differentiable at $x = 1$.

Everywhere else, $f(x)$ is a standard polynomial ($x$ or $x^3$) and thus infinitely differentiable.

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