Is There a Root for cos(x) = x in the Interval (0,1)?

[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).

 

[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 $f$ is continuous on $[a,b]$ and $L$ is strictly between $f(a)$ and $f(b)$, then there is at least one point $c$ in $(a,b)$ satisfying $f(c)=L$.

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 $[0,1]$.

 

[happysmiles36]

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

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.

Yeah, that would have made it much clearer imo.

 

[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.

 

[CellCoree]

I can provide a response to the content regarding the Intermediate Value Theorem. The Intermediate Value Theorem states that for a continuous function f on an interval [a,b], if a value y is between f(a) and f(b), then there exists a value c in the interval [a,b] such that f(c) = y. This means that if a function is continuous on an interval, it will take on every value between its endpoints at some point within that interval.

In the given example, the function cos(x) is continuous on the interval (0,1) and takes on values between 1 and 0.540... at some point within that interval. However, as correctly pointed out, the value 0 is excluded from the domain, so there cannot be a root at x=0. This is because the function is not defined at x=0, and therefore cannot take on the value of 0.

Regarding the statement that cos(x) = x, this is not entirely accurate. As mentioned, the function cos(x) takes on values between 1 and 0.540... within the given interval, but it is not equal to x at any specific point within that interval. The equation cos(x) = x may have a solution at x=~0.79, but this is not the same as saying that the function is equal to x at that point. It simply means that at that point, the function takes on the same value as the independent variable x.

In conclusion, the Intermediate Value Theorem guarantees that there is a root of the given equation in the given interval, but this root cannot be at x=0 due to the exclusion of 0 in the domain. It is important to note that while the function cos(x) and the equation cos(x) = x may seem similar, they are not the same and should not be confused.