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

by jacobrhcp
Tags: work
 P: 169 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 ^_^
Math
Emeritus
Thanks
PF Gold
P: 39,682
 Quote by 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
Here's your error. You proved above that $f'(x)\ge x$. That does NOT give $f'(c)\ge x$, only $f'(c)\ge c$.

 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 ^_^
 P: 169 ah of course. that's great. thanks a lot.

 Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 1 Introductory Physics Homework 9 Introductory Physics Homework 20 Introductory Physics Homework 1