Assume f(0)=f'(0)=0, prove there exists a positive constant such that f(x)>o...

1. The problem statement, all variables and given/known data
Assume that f is a differentiable function such that f(0)=f'(0)=0 and f''(0)>0. Argue that there exists a positive constant a>0 such that f(x)>0 for all x in the interval (0,a). Can anything be concluded about f(x) for negative x's?

2. Relevant equations

3. The attempt at a solution
I think I should use the MVT so here is what I tried:

$$f'(c) = \frac{f(a) - f(0)}{a-0}$$
f(0)=0 is given so:
$$f'(c) = \frac{f(a)}{a}$$

Now I am confused on how to relate this to f(x)>0.
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 Think about what f''(0)>0 means for f'(x) in the immediate vicinity of 0.
 Does it mean that it is increasing? So f'(x)>0? Which means f(x) is increasing from f(0) which means f(x)>0 for a near 0?

Assume f(0)=f'(0)=0, prove there exists a positive constant such that f(x)>o...

... for x in some region just greater than 0, yes.

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 Quote by NWeid1 Does it mean that it is increasing? So f'(x)>0? Which means f(x) is increasing from f(0) which means f(x)>0 for a near 0?
Assuming what you mean by "it is increasing" is "f'(x) is increasing near 0", yes. Good intuition, but of course, that is what you are supposed to prove.

Mentor
 Quote by NWeid1 Does it mean that it is increasing? So f'(x)>0? Which means f(x) is increasing from f(0) which means f(x)>0 for a near 0?
What you say will be clearer if you minimize the number of pronouns such as "it". The question involved f'' and f'. Which one of these do you mean by "it?"
 Ok, I think I got it. If a>0, and f''(0)>0 means f'(0) will be increasing so f'(x)>0 which means f(x) is increasing the origin and therefore f(x)>0. And since a>0 and f(a)>0 f'(c) = f(a)/a > 0 and therefore f(a) > 0. Is this right?

Mentor
 Quote by NWeid1 Ok, I think I got it. If a>0, and f''(0)>0 means f'(0) will be increasing so f'(x)>0
This is not necessarily true. For example, if f(x) = x2, f''(x) > 0 for all x, but the graph of y = f(x) is decreasing over half of its domain.
 Quote by NWeid1 which means f(x) is increasing the origin and therefore f(x)>0. And since a>0 and f(a)>0 f'(c) = f(a)/a > 0 and therefore f(a) > 0. Is this right?

 Quote by Mark44 This is not necessarily true. For example, if f(x) = x2, f''(x) > 0 for all x, but the graph of y = f(x) is decreasing over half of its domain.
Yeah but since we're only looking at x>0, wouldn't it work? So confused, ugh! lol
 Mentor That was just an example. My point is that f''(x) being positive doesn't necessarily mean that f is increasing. If you want an example where x > 0, consider f(x) = (x - 10)2. f''(x) = 2 > 0, but there is an interval, namely [0, 10], on which the graph is decreasing.
 Ok. I got it now. So now I'm confused. I see now that i used f(a)>0 and a>0 to prove that f(a)>0 lol....But, since f''(0)>0, f'(x) will be increasing at 0, right? And if f'(0)=0 and it is increasing, when x>0 for x near 0, f'(x)>0. Is this right at all? lol

Mentor
 Quote by NWeid1 Ok. I got it now. So now I'm confused. I see now that i used f(a)>0 and a>0 to prove that f(a)>0 lol....But, since f''(0)>0, f'(x) will be increasing at 0, right?
f'(x) will be increasing in some interval around 0. Increasing applies to an interval, not just a single point.
 Quote by NWeid1 And if f'(0)=0 and it is increasing, when x>0 for x near 0, f'(x)>0. Is this right at all? lol
Who is "it"?