Answer
$f(b) \lt g(b)$
Work Step by Step
Let $h = f-g$
1. Since $f$ and $g$ are continuous on the interval $[a,b]$, then the function $h$ is continuous on the interval $[a,b]$
2. Since $f$ and $g$ are differentiable on the interval $(a,b)$, then the function $h$ is differentiable on the interval $(a,b)$
$h$ satisfies the hypotheses of the Mean Value Theorem on the interval $[a,b]$
Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(a,b)$ such that $h'(c) = \frac{h(b)-h(a)}{b-a}$
On the interval $(a,b)$:
$h(x) = f(x) - g(x)$
$h'(x) = f'(x)- g'(x)$
Note that $h'(x) \lt 0$ since $f'(x) \lt g'(x)$
Then:
$h'(c) = \frac{h(b)-h(a)}{b-a} \lt 0$
$h(b)-h(a) \lt 0$
$[f(b)-g(b)]-[f(a)-g(a)] \lt 0$
$[f(b)-g(b)]-(0) \lt 0$
$f(b)-g(b) \lt 0$
$f(b) \lt g(b)$
