# Intermediate value theorem

• Sep 30th 2009, 12:49 PM
mjoconn
Intermediate value theorem
Our professor only briefly skimmed over the IVT, and he assigned a problem for homework, and I have no clue what to do.

a) Suppose that F and G are two continuous functions on an interval a < x < b, and that F(a) < G(a) but F(b) > G(b). Show that the equation F(x) = G(x) is satisfied for some x on the interval.

b) By applying a), show it is possible to cut any circular cake through its exact center so that the two halves have exactly the same amount (area) of icing, no matter how unevenly the cake may have been iced.
• Sep 30th 2009, 01:10 PM
TheEmptySet
Quote:

Originally Posted by mjoconn
Our professor only briefly skimmed over the IVT, and he assigned a problem for homework, and I have no clue what to do.

a) Suppose that F and G are two continuous functions on an interval a < x < b, and that F(a) < G(a) but F(b) > G(b). Show that the equation F(x) = G(x) is satisfied for some x on the interval.

b) By applying a), show it is possible to cut any circular cake through its exact center so that the two halves have exactly the same amount (area) of icing, no matter how unevenly the cake may have been iced.

For part a) construct a new function

$\displaystyle u(x)=F(x)-G(x)$

I hope this gets you started. Good luck
Since both F and G are continous so is the difference of the two.

Now notice that

$\displaystyle u(a)=F(a)-G(a)< 0$ and that

$\displaystyle u(b)=F(b)-G(b) >0$

So by the IVT there is $\displaystyle c \in (a,b)$ such that

$\displaystyle u(c)=0$ but

$\displaystyle u(c)=0=F(c)-G(c) \iff F(c)=G(c)$