1. ## prove increasing

f(x) f'(x) f''(x)>0 for all x belonging to R. Prove that f(x) is strictly increasing.

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:

$\displaystyle{0x\,,\,f(y)>f(x)$ 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.

3. Tonio : the limit of positive numbers is not always positive.
If we suppose $f'(x_0)\leq 0$ we have by the mean value theorem $f'<0$ hence $ff''<0$.

4. Originally Posted by girdav
Tonio : the limit of positive numbers is not always positive.

Who said or implied it is? Here we have the other way around: since we know that (1) the limit

exists and (2) the limit is always positive, then we can calculate the limit by taking the limit from the right

and then we get what I wrote.

If we suppose $f'(x_0)\leq 0$ we have by the mean value theorem $f'<0$ hence $ff''<0$.

I really don't understand what you mean to say with the above and how's it related to the OP.

Tonio

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 $x\to-\infty$ to get a contradiction. If f" is negative, think about what happens as $x\to+\infty$, and again get a contradiction.