Please help me with this Calculus question.

Question

Suppose f and g are continuous and differentiable on [a, b], such that f(a) = g(a) and f'x) ≤ g'(x) for x ∈ [a, b]. By using the Mean Value Theorem on the function h = f − g, show f(b) ≤ g(b).

Answer 1

Since f(x) and g(x) are continuous and differentiable on [a,b], so is h(x) = f(x) - g(x).

Also, h'(x) = f'(x) - g'(x) ≤ 0 for all x in [a,b] since f'(x) ≤ g'(x) on the interval [a,b].

By the MVT, there is at least one number,c, in the interval (a,b) so that h'(c) = (h(b) - h(a)) / (b - a).

h'(c) = [f(b) - g(b) - f(a) + g(a)] / (b - a)

Since f(a) = g(a), the equation above simplifies to h'(c) = [f(b) - g(b)] / (b-a) ≤ 0. But, b-a > 0 since a < b.

Therefore, f(b) - g(b) ≤ 0. That is, f(b) ≤ g(b).