# I have something that doesn't work out for me

1. Sep 9, 2007

### jacobrhcp

I'm going to use this theorem of differential analysis: if g(x) is differentiable between a and b, then there is a c for which $$\frac{g(b)-g(a)}{b-a}$$=g'(c)

Let f be differentiable twice, and let f(0)=f'(0)=0 and let f''(x)$$\geq$$1 for all x>0

choose g(x) as f'(x), a = 0, b = x
then there is a c so that:
$$\frac{g(b)-g(a)}{b-a}$$=$$\frac{f'(x)}{x}$$=f''(c)$$\geq$$1 (f''(x) was already bigger then 1)

so f'(x) $$\geq$$ x (for all x > 0)

now I am going to use the theorem again, saying g(x)=f(x), a = 0, b = x

then there is a c so that:

$$\frac{g(b)-g(a)}{b-a}$$=$$\frac{f(x)}{x}$$=f'(c)$$\geq$$x

so f(x) $$\geq$$ x^2 (for all x>0)

but this isn't true. The function 1/2 x^2 also applies for all those things I wanted the function to be, but is smaller than x^2. Anyone knows what I've been doing wrong? This was an old exam exercise, and it's been troubling me since before the holidays.

Ps. this is the first time I did everything with tech on this site, and it's horrible unclear to me if I entered it allright. Hope it worked. If it doesn't, I hope I'll have enough time to edit a bit before someone reads it ^_^

Last edited: Sep 9, 2007
2. Sep 9, 2007

### HallsofIvy

Staff Emeritus
Here's your error. You proved above that $f'(x)\ge x$. That does NOT give $f'(c)\ge x$, only $f'(c)\ge c$.

3. Sep 9, 2007

### jacobrhcp

ah of course. that's great. thanks a lot.