Intermediate Value Theorem

Let $a,b \in \mathbb{R}$ and let $f,g \in C([a,b])$. Suppose $f(a)<g(a)$ and $f(b)<g(b)$. Prove that there is a $c \in (a,b)$ such that $f(c)=g(c)$.

Edit: Solved.

Your thread was originally "Answered", though no longer. It still says "solved" - is it?

As it stands, it's not true. Don't you mean $f(b)>g(b)$ ?

Let $a,b \in \mathbb{R}$ and let $f,g \in C([a,b])$. Suppose $f(a)<g(a)$ and $f(b)>g(b)$. Prove that there is a $c \in (a,b)$ such that $f(c)=g(c)$.

Edit: Solved.

I was thinking of assuming that f(a)<f(b), proving it holds for this case and then prove the other cases (f(b)>f(a), g(a)<g(b), g(b)<g(a)).

For the first case I get:

g(b)<f(b)<f(a)<g(a)

By IVT:
For all y in [g(b),g(a)] there exists c in [a,b] such that g(c)=y.

For all y in [f(b),f(a)] there exists d in [a,b] such that f(d)=z.

Then as f is continuous and bounded below and above by g(b),g(a) respectively, we can set z=y.

So if y=z, this implies, c=d hence, f(c)=g(c).

You don't have to assume f(a)<g(b) - you're told that.

I'm not clear what your cases are there.

You might find a diagram useful.



In bold - interval is [f(a),f(b)], though even the idea that it is an interval requires the IVT.

We can't set z=y, as we don't even know if thest two intervals [f(a),f(b)], [g(b),g(a)] overlap.

The method to use is the one atsruser just posted. Construct the function g(x)-f(x) or the negative thereof, and consider what values it takes at a,b, in particular their sign.

or is the diagram enough to answer the question?

I am not assuming f(a)<g(b), I am assuming f(a)<f(b).

Sorry, misread. The whole thing got to be rather confusing - trying to follow what you were doing.

That's fine. I think I am making it more complicated than it should be.

f(a) - g(a) < 0
f(b) - g(b) > 0

So by IVT there exists a $c$ in (a,b) such that f(c)-g(c)=0.

Then f(c)=g(c).

Is that right?