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

[NWeid1]

Homework Statement

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?

Homework Equations

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.

 

[Joffan]

Think about what f''(0)>0 means for f'(x) in the immediate vicinity of 0.

 

[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?

 

[Joffan]

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

 

[LCKurtz]

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.

 

[Mark44]

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?"

 

[NWeid1]

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?

 

[Mark44]

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.

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?

 

[NWeid1]

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

 

[Mark44]

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.

 

[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? 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

 

[Mark44]

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.

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"?