Big O notation and derivative

Question

Given a function $f(x) = x^2 + g(x)$ such that $g(x) = O(x)$ and that $f'(x)$ is monotonic non-decreasing for all $x \geq x_0$

I need to prove that $g'(x) = O(\sqrt{x})$, and also that without the monotonic condtion such implication is false

For the latter question I have $g(x) = \sin(x^2) = O( x)$ but $g'(x) = 2x \cos(x^2) = O(x)$.

Which means that the solution to the first question $g'(x) = O(\sqrt{x})$ depends on the fact that $f'$ is monotonic but I couldn't prove it. How to prove it then?

Answer 1

Hint. Use the mean value theorem.

The MVT gives, for $x, h > 0$ such that $x-h > x_0$: $$\frac{f(x) - f(x-h)}h \leq f'(x) \leq \frac{f(x+h) - f(x)}h$$ Plugging in $f(x) = x^2+g(x)$: $$\frac{g(x) - g(x-h) - h^2}h \leq g'(x) \leq \frac{g(x+h) - g(x) + h^2}h $$ Now take $h = \sqrt x$ to bound $g'(x)$ from above and below by $O(\sqrt x)$ functions.