Prove the following: Let be real numbers, let and be functions which are differentiable on . Suppose that , that is nonzero on , and exists and equals . Then for all , and exists and equals .
Suppose that for some . Since , then by Rolle's theorem we can find a such that . But this contradicts the fact that is nonzero on . Hence for all .
Now to show that exists and equals , we can show that for any sequence which converges to . How would I show this? I am guessing that I have to create a new function, and somehow manipulate it so that . And this is equivalent to showing that .
The mean value theorem says that there is an x in the interval (a,b) such that .
Now, we know that g' is nonzero (you assumed it), and if g(a) g(b), we can write .
As we know f(a) = g(a) = 0,
If we take the limit of both sides, then x and b will approach the same point.
You could use Rolle's Theorem also right? In terms of sequences you can define a function by . So this is continuous on the interval, and is zero at the endpoints. And . Using Rolle's Theorem we get . as and so as by Squeeze Theorem.
And so .
But what you did here was proving Mean Value Theorem using Rolle's Theorem =)
Originally Posted by particlejohn
(except you used sequences instead of function. see here)