f(x) = 0 for 0 ≤ x < 1 = x for 1 ≤ x < 2 = 2(x – 1) for 2 ≤ x < 3 ∴ For 0 ≤ x < 1, f o f(x) = f[f(x)] = f(0) = 0. Now let 1 ≤ x < 2 then f o f(x) = f[f(x)] = f(x) = x Now let 2 ≤ x < 5/2 then 2 ≤ 2x – 2 < 3 f o f(x) = f[f(x)] = f(2x – 2) = 2(2x – 2) – 2 = 4x – 6. and if x ≥ 5/2 , 2x – 2 ≥ 3, so f(x) is undefined. Hence g(x) = f o f(x) = 0 for 0 ≤ x < 1 = x for 1 ≤ x < 2 = 4x – 6 for 2 ≤ x < 5/2 .Domain of f o f(x) is [0, 5/2). Observe that g(x) is continuous but not differentiable at x = 2. Now g′(x) = 1 > 0 for 1 ≤ x < 2 = 4 > 0 for 2 ≤ x < 5/2 g(2) = 2 ∴ g(x) has no max or min at 2