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clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Intersecting Functions?

if f and g are continuous functions on [0,1] such that
f(0)<g(0) and f(1)>g(1). Show that there must be an x in (0,1) for which f(x)=g(x)...

I know it's true and can prove it geometrically... but i think i should be proving it using calculus. I should probably be using the intermediate value theorem right?

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Intersecting Functions?

Yep, this is a job for the good ol' IVT.

Try considering h(x) := f(x) - g(x).

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Why did the vector cross the road?
It wanted to be normal.

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Re: Intersecting Functions?

Alright, I have h(x) = f(x) - g(x).
By def. of IVT (from my book) it says "Given h(x) on interval [a,b] and any x1 and x2 in the interval; then for any number, N, such that h(x1)<N<h(x2) then there exists as c in (a,b) such that h(c)=N."

So I have h(x) on the interval [0,1] with x1=0 and x2=1. So for any N between h(0) and h(1) there is a c such that h(c)=N.
So what new do i know?  h(0)<=f(0)<=g(0) and g(1)<h(1)<f(1)...
So what? I'm not sure how this exactly helps.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Intersecting Functions?

You're off to a good start, but don't compare h(0) and h(1) to other functions.
Instead, think about what sign their values will be, and use the IVT around that.

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Why did the vector cross the road?
It wanted to be normal.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Intersecting Functions?

A specific use of the IVT is to approximate 0's of a function.  This is because if f is continuous, f(a) < 0 and f(b) > 0, then there must exist a point c in [a, b] where f(c) = 0.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

clooneyisagenius

Member

Registered: 2007-03-25

Posts: 56

Re: Intersecting Functions?

Gotcha mathsyperson thanks!

So because h(0) is less than 0 and h(1) is greater than 0 then at some point in the interval of (0,1) we will have a value,c, so that h(c) = 0. And thus, since h(x)=f(x)-g(x) this point is shared by f(x) and g(x)!

Thank you!!!