proof about derivatives of functions

Question

Suppose the functions f and g are continuous on [a, b] and differentiable on (a, b).

Assume that $f(a) = g(a)$.

Prove that if $f'(x) <g'(x)$ for all $x ∈ (a, b),$ then $f(b) < g(b)$.

To me, intuitively, I thought that this would not be true however I could not find a counter example, so now I am a bit lost on how to prove this.

Answer 1

Define $h : [a, b] \to \mathbb{R}$ as $h(x) = g(x) - f(x)$ and apply the Mean Value Theorem to $h$ on $[a, b]$ to get a $\xi \in (a, b)$ with

$$h(b) - h(a) = (b-a) h'(\xi).$$

Then

$$g(b) - f(b) = h(b) - h(a) = (b-a) \big( g'(\xi) - f'(\xi) \big) > 0,$$

hence $f(b) < g(b)$.

Answer 2

$g'-f'$ is positive so that $g-f$ is growing from $0$, hence remains positive.