# Thread: Proof relating to Rolle's and MVT

1. ## 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!

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

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

3. 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!
Originally Posted by NonCommAlg
[this is false if $\displaystyle x < 0.$ for example $\displaystyle f(x)=2x, \ g(x)=x.$] anyway, for $\displaystyle x > 0,$ let $\displaystyle h(x)=f(x)-g(x).$ then $\displaystyle h(0)=0$ and by MVT there exists $\displaystyle 0 < c < x$ such that:

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