Let 𝑓: ℝ → ℝ be a function such that 𝑓(𝑥) = max{𝑥, \(x^3\) }, 𝑥 ∈ ℝ ,…

Key idea: f(x) = max{x, x^3}. Nondifferentiability can occur only where the two branches meet, i.e., where x = x^3.

• Solve x = x^3: x(x-1)(x+1)=0, so the intersection points are x = −1, 0, 1.

• Derivatives of the branches: d/dx(x) = 1 and d/dx(x^3) = 3x^2.

• At x = −1: for x<−1 we have x > x^3 so f(x)=x and the left derivative = 1; for x>−1 we have x^3 > x so f(x)=x^3 and the right derivative = 3·(−1)^2 = 3. Since 1 ≠ 3, f is not differentiable at −1.

• At x = 0: for x<0 we have x^3 > x so f(x)=x^3 and the left derivative = 3·0^2 = 0; for x>0 we have x > x^3 so f(x)=x and the right derivative = 1. Since 0 ≠ 1, f is not differentiable at 0.

• At x = 1: for x<1 we have x > x^3 so f(x)=x and the left derivative = 1; for x>1 we have x^3 > x so f(x)=x^3 and the right derivative = 3·1^2 = 3. Since 1 ≠ 3, f is not differentiable at 1.

Conclusion: The function f is not differentiable exactly at the intersection points {−1, 0, 1}.