) Let f, g be differentiable functions on an interval I. Suppose that a, b ∈ Iwith f (b) ≥ g(b) and f ′(x) ≤ g′(x) on [a, b]. Show that f (x) ≥ g(x) on[a, b].

To prove that f(x) ≥ g(x) on [a, b], we can use the Mean Value Theorem.

First, let's consider the function h(x) = f(x) - g(x). We want to show that h(x) ≥ 0 on [a, b].

Since f(x) and g(x) are differentiable on [a, b], h(x) is also differentiable on [a, b].

Now, let's consider the derivative of h(x), denoted as h'(x). Using the properties of derivatives, we have:

h'(x) = f'(x) - g'(x)

Given that f'(x) ≤ g'(x) on [a, b], we can conclude that h'(x) ≤ 0 on [a, b].

According to the Mean Value Theorem, if a function is differentiable on an interval and its derivative is non-positive on that interval, then the function is non-increasing on that interval.

Since h'(x) ≤ 0 on [a, b], it follows that h(x) is non-increasing on [a, b].

Now, let's consider the values of h(x) at the endpoints of the interval. We have:

h(a) = f(a) - g(a) ≥ 0 (since f(b) ≥ g(b)) h(b) = f(b) - g(b) ≥ 0

Since h(x) is non-increasing on [a, b] and h(a) ≥ 0, it follows that h(x) ≥ 0 for all x in [a, b].

Therefore, we can conclude that f(x) ≥ g(x) on [a, b].