Question

Let a function \(g : [0, 4] \to R\) be defined as
\(g(x) = \begin{cases} \max\limits_{0 \le t \le x} \{t^3 - 6t^2 + 9t - 3),\quad 0 \le x \le 3 \\ 4 - x,\quad 3 \lt x \le 4 \end{cases}\)
, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ______________.

Hint:

Let \(f(t) = t^3 - 6t^2 + 9t - 3\); f rises on [0,1] and falls on [1,3], so the running maximum equals f(x) up to x = 1 and stays at f(1) = 1 afterwards.

Answer 1

Correct Answer: 1

\(f(t) = t^3 - 6t^2 + 9t - 3\), \(f'(t) = 3(t-1)(t-3)\). f increases on [0, 1] and decreases on [1, 3]; f(0) = -3, f(1) = 1, f(3) = -3.
Running maximum on [0, x]: for \(0 \le x \le 1\) it is f(x); for \(1 \le x \le 3\) it stays 1 (f only decreases after 1 and never exceeds 1).
So \(g(x) = f(x)\) on [0,1], \(g(x) = 1\) on [1,3], \(g(x) = 4 - x\) on (3,4].
At x = 1: left derivative \(f'(1) = 0\), right derivative 0, so differentiable. At x = 3: g(3) = 1 and \(4 - 3 = 1\) (continuous), left derivative 0, right derivative -1, so NOT differentiable.
Only x = 3 in (0, 4) fails differentiability.
Answer: 1