# Intermediate value theorem proof

• Aug 13th 2011, 09:47 PM
whiteboard
Intermediate value theorem proof
have a continuous function f:[0,1] -> [0,1]

For some $x1, f(x1)>x1$
For some $x2, f(x2)>x2$

Using the Intermediate Value Theorem, show that there exists some x* such that f(x*)=x*

I have attempted the problem:

Make $h(x)=f(x)-x$. Then we are looking for the case when $h(x)=0$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that $f(0)>0$ because f must take on a value in the interval [0,1], and the same for $f(1)<1$. These two statements imply that $h(0)>0 and h(1)<0$. So then by the intermediate value theorem, we can say that $h(x)=0$ for some $x [0,1]$ which means that $f(x)=x$ for some $x [0,1]$

As I am pretty rusty on my proof skills I was wondering if somebody could check this for me.

Thanks
• Aug 13th 2011, 10:17 PM
CaptainBlack
Re: Intermediate value theorem proof
Quote:

Originally Posted by whiteboard
have a continuous function f:[0,1] -> [0,1]

For some $x1, f(x1){\color{red}<}x1$
For some $x2, f(x2)>x2$

Using the Intermediate Value Theorem, show that there exists some x* such that f(x*)=x*

I have attempted the problem:

Make $h(x)=f(x)-x$. Then we are looking for the case when $h(x)=0$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that $f(0)>0$ because f must take on a value in the interval [0,1], and the same for $f(1)<1$. These two statements imply that $h(0)>0 and h(1)<0$. So then by the intermediate value theorem, we can say that $h(x)=0$ for some $x [0,1]$ which means that $f(x)=x$ for some $x [0,1]$

As I am pretty rusty on my proof skills I was wondering if somebody could check this for me.

Thanks

The function $h(x)=f(x)-x$ is continuous on an interval containing both $x_1$ and $x_2$ , with $h(x_1)<0$ and $h(x_2)>0$, hence by theIntermediate value theorem there exists a $c$ between $x_1$ and $x_2$ such that $h(c)=0$ and so $f(c)=c$

CB