Question

Let f(x) = x3 – x2 + x + 1 and g(x) = {(max.{f(t) : 0 ≤ t ≤ x} : 0 ≤ x ≤ 1), (3 - x, : 1 < x ≤ 2) Discuss the continuity and differentiability of the function g(x) in (0, 2).

Answer 1

Given f(x) = x3 – x2 + x + 1

f'(x) = 3x2 – 2x + 1

Since its D is negative,

so, f'(x) > 0, ∀ x ∈R

Thus, f is strictly increasing in (0, 2)

Clearly, g(x) is continuous in (0, 2) but not differentiable at x = 1