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

Re: Intermediate value theorem proof

Quote:

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**whiteboard** 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