Proving a statement involving the intermediate value theorem

• Aug 6th 2010, 11:27 PM
Glitch
Proving a statement involving the intermediate value theorem
The question:

Suppose that f is continuous on [0,1] and that Range(f) is a subset of [0,1]. By using g(x) = f(x) - x, prove that there is a real number c in [0,1] such that f(c) = c.

I'm not sure where to start. I know that the intermediate value theorem applies, since we're given a function g(x) which is continuous on the interval [0,1]. However, I don't know how to use this to prove that f(c) = c.

Any help would be great.
• Aug 7th 2010, 02:56 AM
Plato
What can be said about the values of \$\displaystyle g(0)~\&~g(1)?\$.

Recall that you are given \$\displaystyle 0\le f(x) \le 1\$.
• Aug 7th 2010, 03:52 AM
Glitch
I'm not sure what can be said. I don't know what f(x) is, I just know part of its domain and range. I'm also not sure how we are given \$\displaystyle 0\le f(x) \le 0\$
• Aug 7th 2010, 04:15 AM
HallsofIvy
That was a typo. Plato meant to say \$\displaystyle 0\le f(x)\le 1\$. (That may be the first mistake he has ever made!)

First, if either f(0)=0 or f(1)= 1, we are done. So we can assume that f(0)> 0 and that f(1)< 1.

Let g(x)= f(x)- x as you say. Then g(0)= f(0). Is that positive or negative?

g(1)= f(1)- 1. Is that positive or negative?
• Aug 7th 2010, 05:25 AM
Glitch
g(0) would be positive, g(1) would be negative.
• Aug 7th 2010, 05:37 AM
Plato
Quote:

Originally Posted by Glitch
g(0) would be positive, g(1) would be negative.

Correct. Now what?
• Aug 7th 2010, 05:43 AM
Glitch
Hmm, I'm not sure. I feel like I'm missing something obvious here...
• Aug 7th 2010, 05:48 AM
Plato
\$\displaystyle g(0)>0~\&~g(1)<0\$.

Because \$\displaystyle g\$ is continuous apply the MVT.