Results 1 to 2 of 2

Thread: Intermediate value theorem proof

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    7

    Intermediate value theorem proof

    have a continuous function f:[0,1] -> [0,1]

    For some$\displaystyle x1, f(x1)>x1 $
    For some$\displaystyle 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 $\displaystyle h(x)=f(x)-x$. Then we are looking for the case when $\displaystyle h(x)=0$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that $\displaystyle f(0)>0$ because f must take on a value in the interval [0,1], and the same for $\displaystyle f(1)<1$. These two statements imply that $\displaystyle h(0)>0 and h(1)<0$. So then by the intermediate value theorem, we can say that $\displaystyle h(x)=0$ for some $\displaystyle x [0,1] $ which means that $\displaystyle f(x)=x $ for some $\displaystyle x [0,1] $

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

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5

    Re: Intermediate value theorem proof

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

    For some$\displaystyle x1, f(x1){\color{red}<}x1 $
    For some$\displaystyle 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 $\displaystyle h(x)=f(x)-x$. Then we are looking for the case when $\displaystyle h(x)=0$. So if f(0)=0 or f(1)=1, then we are done. So we assume neither case is true. Then we have that $\displaystyle f(0)>0$ because f must take on a value in the interval [0,1], and the same for $\displaystyle f(1)<1$. These two statements imply that $\displaystyle h(0)>0 and h(1)<0$. So then by the intermediate value theorem, we can say that $\displaystyle h(x)=0$ for some $\displaystyle x [0,1] $ which means that $\displaystyle f(x)=x $ for some $\displaystyle 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 $\displaystyle h(x)=f(x)-x$ is continuous on an interval containing both $\displaystyle x_1$ and $\displaystyle x_2$ , with $\displaystyle h(x_1)<0$ and $\displaystyle h(x_2)>0$, hence by theIntermediate value theorem there exists a $\displaystyle c$ between $\displaystyle x_1$ and $\displaystyle x_2$ such that $\displaystyle h(c)=0$ and so $\displaystyle f(c)=c$

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Possibly an intermediate value theorem proof, but I'm not sure.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Jul 4th 2011, 09:08 AM
  2. Proof of intermediate value theorem for a particular case
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: Apr 26th 2011, 06:37 AM
  3. intermediate value theorem/rolle's theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Dec 8th 2007, 01:55 PM
  4. Proof with Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 28th 2007, 07:53 PM
  5. Proof using the Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 11th 2006, 05:55 PM

/mathhelpforum @mathhelpforum