# Proof relating to Rolle's and MVT

• Dec 13th 2008, 04:57 PM
lakesfan210
Proof relating to Rolle's and MVT
Hello all,

I am being asked to prove that if f(0)=g(0) and f'>g', then f(x)>g(x). It makes perfect intuitive sense, but I am having difficulties in proving it rigorously. Any suggestions anyone could offer to point me in the right direction? Thank you very much for your help!
• Dec 13th 2008, 05:27 PM
NonCommAlg
Quote:

Originally Posted by lakesfan210
Hello all,

I am being asked to prove that if f(0)=g(0) and f'>g', then f(x)>g(x). It makes perfect intuitive sense, but I am having difficulties in proving it rigorously. Any suggestions anyone could offer to point me in the right direction? Thank you very much for your help!

[this is false if $x < 0.$ for example $f(x)=2x, \ g(x)=x.$] anyway, for $x > 0,$ let $h(x)=f(x)-g(x).$ then $h(0)=0$ and by MVT there exists $0 < c < x$ such that:

$f(x)-g(x)=h(x)-h(0)=xh'(c)=x(f'(c)-g'(c)) > 0.$
• Dec 13th 2008, 06:47 PM
Mathstud28
Quote:

Originally Posted by lakesfan210
Hello all,

I am being asked to prove that if f(0)=g(0) and f'>g', then f(x)>g(x). It makes perfect intuitive sense, but I am having difficulties in proving it rigorously. Any suggestions anyone could offer to point me in the right direction? Thank you very much for your help!

Quote:

Originally Posted by NonCommAlg
[this is false if $x < 0.$ for example $f(x)=2x, \ g(x)=x.$] anyway, for $x > 0,$ let $h(x)=f(x)-g(x).$ then $h(0)=0$ and by MVT there exists $0 < c < x$ such that:

$f(x)-g(x)=h(x)-h(0)=xh'(c)=x(f'(c)-g'(c)) > 0.$

Alternatively consider the interval $[0,\xi]$ it is apparent that there exists a $c,c_1\in(0,\xi)$ such that $f'(c)=\frac{f(\xi)-f(0)}{\xi-0}=\frac{f(\xi)-f(0)}{\xi}$ and $g'(c_1)=\frac{g(\xi)-g(0)}{\xi-0}=\frac{g(\xi)-g(0)}{\xi}$. So in any interval $[0,\xi]$ we have that there is a point such that $\frac{g(\xi)-g(0)}{\xi}<\frac{f(\xi)-f(0)}{\xi}\implies g(\xi) and since $\xi$ was arbitrary this completes the proof.