# Thread: Prove using (ONLY) Rolle's Theorem (??)

1. ## Prove using (ONLY) Rolle's Theorem (??)

How do I prove the following theorem, using ONLY Rolle's theorem?

Suppose that for some real number a,
f(x) is continuous on [a, inf)
and f '(x) exists on (a,inf)
with f '(x)>0 for all x>a
Then f(x)>f(a) for all x>a

I think I'm right in saying that Rolle's theorem is;

if a function f of real value is continuous on [a,b], differentiable on (a,b) and f(a)=f(b) then there exists a C in (a,b) such that f'(C)=0

but how do i PROVE the above theorem from that?

Thank you

2. Originally Posted by DBsHo
How do I prove the following theorem, using ONLY Rolle's theorem?

Suppose that for some real number a,
f(x) is continuous on [a, inf)
and f '(x) exists on (a,inf)
with f '(x)>0 for all x>a
Then f(x)>f(a) for all x>a

I think I'm right in saying that Rolle's theorem is;

if a function f of real value is continuous on [a,b], differentiable on (a,b) and f(a)=f(b) then there exists a C in (a,b) such that f'(C)=0

but how do i PROVE the above theorem from that?

Clearly, we must have that either $\displaystyle f(x)<f(a)$ or $\displaystyle f(x)>f(a)$ for all $\displaystyle x\in[a,\infty]$. To see this suppose that there were some points where $\displaystyle f(x)<f(a)$ and some where $\displaystyle f(x)>f(a)$. Then, since $\displaystyle f$ is continuous the IVT furnishes us with a $\displaystyle \xi\in[a,\infty]$ such that $\displaystyle f\left(\xi\right)=f(a)$ at which point Rolle's theorem tells us that $\displaystyle f'(x)=0$ for some point in $\displaystyle (a,\infty)$, this is a contradiction.
4. Basically he has shown that you must have that either $\displaystyle f(x)>f(a)$ or $\displaystyle f(x)<f(a)$ for all $\displaystyle x\in (a,\infty)$. Otherwise you would contradict $\displaystyle f'(x)>0$ $\displaystyle \forall x\in (a,\infty)$. Then since $\displaystyle f'(x)>$ you know which of the choices must be correct.