Real analysis Help - Intermediate Value Thrm

[CrazyCalcGirl]

Homework Statement

Use the Intermediate Value Theorem to show that the equation 2^x=3x has a solution c element of (0,1)

Homework Equations

The Attempt at a Solution

Ok I know this theorem is usually very easy, but I've never done one where I couldn't easily solve for x and plug in the end points to look for a sign change. I already graphed 2^x and 3x on my calculator and found that there is a solution in (0,1) now I just need to prove it.

My only thoughts were this

let f(x)= 2^x
Let g(x) = 3x

f(0)=2^0=1 and f(1)=2^1=2
g(0)=3x0=0 and G(1)= 3x1=3

f(0)>g(0) and f(1)<g(1) so this means at some point they must cross each other and there must exist c element of (0,1) such that 2^x=3x.

Is this even right at all or anyone have a better idea?

 

[ZioX]

You do this trick a lot in analysis when considering two functions being equal at a point or on an interval:

Define
$$f(x):=2^x-3x$$

What do you know about the sum of two continuous functions? Or rather, the sum of a continuous function and a multiple of a continuous function?

 

[CrazyCalcGirl]

I know the sum of two continuous functions is continuous and same for a multiple, but I don't see how this applies to me using the IVT. The problems requires I use that theorem.

If you think the way I did it is wrong how about this...?

If I use your idea to let f(x)= 2^x-3x now I can just plug in the endpoints of my interval.

f(0)=1
f(1)=-1

Since they have opposite signs the IVT says there must be a real root c in (0,1).

We just got into continuity and we haven't actually learned anything about sums or multiples yet. We are not allowed to use anything on the homework we have not done in class.

 

[Gib Z]

By Plugging in the endpoints, you are correct =]

 

[CrazyCalcGirl]

hmm both ways I did it I plugged in the endpoints.. so if the first way correct that I originally posted or is the second way correct where you set f(x)= 2^x-3x?

Thanks : )

 

[Gib Z]

If I use your idea to let f(x)= 2^x-3x now I can just plug in the endpoints of my interval.

f(0)=1
f(1)=-1

Since they have opposite signs the IVT says there must be a real root c in (0,1).

That is the correct part.