Chapter 4 - Applications of the Derivative - 4.3 The Mean Value Theorem and Monotonicity - Exercises - Page 190: 64

$f(x)-g(x) $ is non-increasing.

Consider $$y= f(x)-g(x) $$ Then $$y'= f'(x)-g'(x) $$ Since $f (x) \geq g(x)$ for all $x \geq 0$, then $y' \leq 0$ for all $x \geq 0$. Hence, $y$ is non-increasing. Since $y(0) = f (0) − g(0) = 0$, then $y(x) \leq 0 $ for all $x \geq 0$. Hence, $f (x) − g(x) \leq 0$ for all $x \geq0$, and $f (x) \leq g(x)$ for $ x \geq 0$.