Connection between uniform continuity and derivatives' behaviour

Question

I have a function $f(x)$ continuous on $[0; +\infty)$ and also first and second its derivatives are continuous on the same ray $[0; +\infty)$. So I need to prove:

1) If $f'(x) > 0$ and $f''(x) < 0$ than $f(x)$ is uniformly continuous on $[0;+\infty)$

2) If $f'(x) < 0$ and $f''(x) > 0$ than $f(x)$ is uniformly continuous on $[0;+\infty)$

I understand it intuitively because 1) is something like $\sqrt{x}$ which is uniformly continuous on $[0; +\infty)$ and 2) is smth like $\frac{1}{x+1}$. But how to prove it rigorously? $\epsilon-\delta$ language? Would be grateful for any hints.

Answer 1

For 1) For $x>y\geq 0$, \begin{align*} f(x)-f(y)=f'(\xi_{x,y})(x-y), \end{align*} where $y<\xi_{x,y}<x$, and \begin{align*} f'(\xi_{x,y})-f'(0)=f''(\eta_{x,y})(\xi_{x,y})<0, \end{align*} where $0<\eta_{x,y}<\xi_{x,y}$, so \begin{align*} f(x)-f(y)<f'(0)(x-y). \end{align*} For $\epsilon>0$, for all $x,y\in[0,\infty)$ with $|x-y|<\dfrac{\epsilon}{f'(0)}$, then $|f(x)-f(y)|<\dfrac{f'(0)}{f'(0)}\epsilon=\epsilon$.

Answer 2

A function is uniformly continuous if it has a bounded derivative since by the MVT, $$|f(x) - f(y)| = |f'(\xi)||x-y|\leqslant \sup_{z} |f'(z)| |x-y|.$$

For (1) , f' is continuous, decreasing and bounded below by $0$. Thus $f'(0) \geqslant f'(x) > 0$. Similarly for (2).