If x ∈ [–1, 1], f(g(x)) = sin x – 1 for –1 ≤ x < 0 = sin2 x for 0 ≤ x ≤ 1 |g(x)| = –sin x for –1 ≤ x < 0 = sin x for 0 ≤ x ≤ 1 f(|g(x)|) = sin2 x ∀ x ∈ [–1, 1]. Further |(g(x))| = 1 – sin x for –1 ≤ x < 0 = sin2 x for 0 ≤ x ≤ 1. ∴ h(x) = sin2 x – sin x + 1 for –1 ≤ x < 0 = 2sin2 x for 0 ≤ x ≤ 1 h′(x) = sin 2x – cos x for –1 < x < 0 = 4sin x cos x for 0 < x < 1. The value of h(x) at x =0 is less than the neighboring values. So local minimum at x = 0