f(x) f'(x) f''(x)>0 for all x belonging to R. Prove that f(x) is strictly increasing.
Since thederivative of f at any point exists, we can evaluate it by taking the right limit:
$\displaystyle \displaystyle{0<f'(x)=\lim_{y\to x^+}\frac{f(y)-f(x)}{y-x}\Longrightarrow \forall\,y>x\,,\,f(y)>f(x)$ since the
expression in the fraction has positive denominator and the whole fraction is positive...
Tonio
Ps. Of course, we don't need f''(x) at all...not even its existence.
You may find some of the ideas in this post useful. In particular, it follows from the intermediate value theorem and Darboux's theorem that none of the functions f, f', f" can ever change sign. You want to prove that f' is always positive. So suppose (in order to get a contradiction) that f' is always negative. Since the product ff'f" is positive, it follows that either f or f" must be negative. If f is negative, think about what happens as $\displaystyle x\to-\infty$ to get a contradiction. If f" is negative, think about what happens as $\displaystyle x\to+\infty$, and again get a contradiction.