9367 54 proof.

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

9367 54 proof.

Postby yoyobarn » Mon Dec 19, 2011 12:17 pm

Hi,

How do we prove the answer for 9367 Q54?

If f and g are real-valued differentiable functions and if f'(x)>=g'(x) for all x in the closed interval [0,1], which of the following must be true?

(C) f(1)-g(1)>=f(0)-g(0)

Can we prove it by
\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\geq\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}

\therefore f(x)-f(0)\geq g(x)-g(0), for all x in [0,1]

\therefore f(1)-f(0)\geq g(1)-g(0)

Thanks.

echo
Posts: 18
Joined: Fri Dec 16, 2011 7:41 pm

Re: 9367 54 proof.

Postby echo » Mon Dec 19, 2011 1:31 pm

Recall that by the FTC:
f(1) = f(0) + \int_0^1 f'(x)dx,\,g(1) = g(0)+\int_0^1 g'(x)dx
That should provide the necessary ingredients to finish the proof.

Your proof is invalid because it only applies to x sufficiently close to zero.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 54 proof.

Postby yoyobarn » Mon Dec 19, 2011 1:33 pm

Thanks!!

I<3 Gauss
Posts: 2
Joined: Mon Dec 19, 2011 7:38 pm

Re: 9367 54 proof.

Postby I<3 Gauss » Mon Dec 19, 2011 7:46 pm

Hmm, an alternate proof from the one above is letting h(x)=f(x)-g(x). Since differentiation is linear, h'(x) = f'(x) - g'(x). We also see that h'(x)>=0 since f'(x)>=g'(x). By the Mean Value Theorem, there exists a c in [0,1] such that h'(c) = h(1) - h(0), and h(1) - h(0)>=0. Thus, we have f(1) - g(1) - f(0) + g(0)>=0, or f(1) - f(0) >= g(1) - g(0).




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