Let a function g : [ 0, 4 ] → R be defined as
g(x) = { max _(0 ≤ t ≤ x) { t³ - 6t² + 9t - 3),, 0 ≤ x ≤ 3; 4 - x,, 3 < x ≤ 4 , then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.

Solution

Maths sir · 5 steps

Find where the running maximum stops growing, then compare the slopes at each join.

1. Shape of f. Let f(t) = t³ - 6t² + 9t - 3. Then f'(t) = 3t² - 12t + 9 = 3(t-1)(t-3), so f rises on [0, 1] and falls on [1, 3]. The values are f(0) = -3, f(1) = 1 and f(3) = -3.
2. Running maximum. For 0 ≤ x ≤ 1, f is still rising, so the largest value on [0, x] is f(x) itself. For 1 ≤ x ≤ 3, f only falls after its peak, so the maximum stays fixed at f(1) = 1.

   g(x) = x³ - 6x² + 9x - 3, 0 ≤ x ≤ 1 ; 1, 1 < x ≤ 3 ; 4 - x, 3 < x ≤ 4

3. Check x = 1. The left derivative is f'(1) = 0 and the right derivative, from the constant piece, is also 0. The slopes match, so g is differentiable at x = 1.
4. Check x = 3. Both sides give the value 1, so g is continuous at x = 3. But the left derivative is 0 and the right derivative is -1, so there is a corner.
5. So. Every piece is a polynomial, so it is differentiable inside its own interval. The only point in (0, 4) where g is not differentiable is x = 3, so the answer is 1.

The common slip: 2. Treating x = 1 as a corner gives 2. It is not one, because f'(1) = 0 matches the zero slope of the flat piece.

The idea to keep. A running maximum follows the function while it rises and goes flat once it starts falling. Then check the left and right slopes at every point where the formula changes.

Answer from the NTA final answer key. Solution by PrepWiser.