# Intermediate Value Theorem

1. Sep 13, 2014

### happysmiles36

From the Intermediate Value Theorem, is it guaranteed that there is a root of the given equation in the given interval?

cos(x)=x, (0,1)

cos(0)= 1
cox(1)= 0.540...

So using intermediate value theorem, no.

and

x=0 can't be possible because 0 was excluded in the domain by the round bracket so 0<x<1. Therefore there can't be a root. In other words the domain makes it so the graph is discontinuous at 0 (f(0) does not exist) and if the graph/function doesn't at that point there cannot be a root there.

The answer is yes, but I am struggling to understand why and I am not sure if the books answer is wrong or if I'm wrong.

*I also don't know why cos(x) = x , because when x=1 they aren't equal and would only be true at 1 point in this domain (when x=~0.79).

Last edited: Sep 13, 2014
2. Sep 13, 2014

### gopher_p

It's probably better to examine the equivalent question; Is there a solution to $\cos x-x=0$ on the interval $(0,1)$?

The Intermediate Value Theorem says
If we put $f(x)=\cos x-x$, your (revised) question is asking whether there is $c$ in $(0,1)$ satisfying $f(c)=0$, right?

Given that $f(0)=\cos 0-0=1>0$ and $f(1)=\cos 1-1<0$ (you should check that), do you see how the IVT guarantees a $c$ in $(0,1)$ with $f(c)=0$? Do you see how this is the same as saying $\cos c=c$?

Note that you also need to say something regarding the continuity of $f(x)=\cos x-x$ on the closed interval $[0,1]$.

3. Sep 13, 2014

### happysmiles36

Oh ok, that makes sense, I thought the question was asking me to do something else.

Yeah, that would have made it much clearer imo.

Last edited: Sep 13, 2014
4. Sep 13, 2014

### gopher_p

For what it's worth, I'm not a big fan of the use of the word "root" in the problem statement. It'd probably be better to use the word "solution" in that context.