1. ## Continuous Functions

How would I go about doing this question:

$\displaystyle Suppose\ that\ f\ and\ g$ $\displaystyle are\ continuous\ functions\ on\ the\ closed\ interval$ $\displaystyle [0,1]\ and\ that\ 0 \leq f(x) \leq 1\ for\ every\ x\ in\ [0,1].$ $\displaystyle Show\ that\ there\ exists\ a\ real\ number$ $\displaystyle c \in [0,1]\ such\ that\ f(c)=c.$

Would you do this question using the intermediate value theorem to $\displaystyle g(x)=f(x)-x$

2. Originally Posted by acevipa
How would I go about doing this question:

$\displaystyle Suppose\ that\ f\ and\ g$ $\displaystyle are\ continuous\ functions\ on\ the\ closed\ interval$ $\displaystyle [0,1]\ and\ that\ 0 \leq f(x) \leq 1\ for\ every\ x\ in\ [0,1].$ $\displaystyle Show\ that\ there\ exists\ a\ real\ number$ $\displaystyle c \in [0,1]\ such\ that\ f(c)=c.$

Would you do this question using the intermediate value theorem to $\displaystyle g(x)=f(x)-x$
Yes, that's a good idea. I suggest you look at g(0) and g(1).

3. Originally Posted by HallsofIvy
Yes, that's a good idea. I suggest you look at g(0) and g(1).
Would you do it like this

$\displaystyle \frac{g(1)-g(0)}{1-0} = g'(c)$

$\displaystyle \frac{f(1)-1-f(0)+0}{1-0}=f'(c)-1$

$\displaystyle Given\ that\ f(1)=1\ and\ f(0)=0$

$\displaystyle \frac{1-1-0}{1}=f'(c)-1$

$\displaystyle \frac{0}{1}=f'(c)-1$

$\displaystyle f'(c)-1=0$

$\displaystyle f'(c)=1\ \forall c \in (0,1)$

What do I do from here?