Let f, g : R → R be differentiable on [1, 2]. Suppose that f (1) = g(1) − 1 andf ′(x) ≤ g′(x) − 1 for all x ∈ [1, 2].Prove thatg(2) ≥ f (2) + 2.Hint) Consider the function h(x) := g(x) − x − f (x). Prove that h is non-negative on [1, 2

To prove the given inequality, we will use the hint and consider the function h(x) := g(x) - x - f(x).

Step 1: Define the function h(x) Let's define a function h(x) = g(x) - x - f(x).

Step 2: Find h(1) We know that f(1) = g(1) - 1, so h(1) = g(1) - 1 - 1 - f(1) = 0.

Step 3: Find h'(x) The derivative of h(x) is h'(x) = g'(x) - 1 - f'(x).

Step 4: Use the given inequality We are given that f'(x) ≤ g'(x) - 1 for all x in [1, 2]. This implies that -f'(x) ≥ 1 - g'(x). So, h'(x) = g'(x) - 1 - f'(x) ≥ 0.

Step 5: Apply the Mean Value Theorem Since h(x) is differentiable on [1, 2], by the Mean Value Theorem, there exists a c in (1, 2) such that h'(c) = (h(2) - h(1)) / (2 - 1). Since h'(x) ≥ 0, we have h(2) - h(1) ≥ 0, which implies that h(2) ≥ h(1).

Step 6: Conclude the proof Since h(1) = 0, we have h(2) ≥ 0. But h(2) = g(2) - 2 - f(2), so g(2) - f(2) ≥ 2, which is the inequality we wanted to prove.