therightwomanwf

2022-07-12

What steps would I take or use in order to use the intermediate value theorem to show that $\mathrm{cos}x=x$ has a solution between $x=0$ and $x=1$?

Asdrubali2r

EDIT

Recall the statement of the intermediate value theorem.

Theorem If $f\left(x\right)$ is a real-valued continuous function on the interval [a,b], then given any $y\in \left[min\left(f\left(a\right),f\left(b\right)\right),max\left(f\left(a\right),f\left(b\right)\right)\right]$, there exists $c\in \left[a,b\right]$ such that $f\left(c\right)=y$.

The theorem guarantees us that given any value $y$ in-between $f\left(a\right)$ and $f\left(b\right)$, the continuous function $f\left(x\right)$ takes the value $y$ for some point in the interval [a,b].

Now lets get back to our problem. Look at the function $f\left(x\right)=\mathrm{cos}\left(x\right)-x$.

We have $f\left(0\right)=1>0$.

We also have that $f\left(1\right)=\mathrm{cos}\left(1\right)-1$. But $\mathrm{cos}\left(x\right)<1$, $\mathrm{\forall }x\ne 2n\pi$, where $n\in \mathbb{Z}$. Clearly, $1\ne 2n\pi$, where $n\in \mathbb{Z}$. Hence, we have that $\mathrm{cos}\left(1\right)<1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}f\left(1\right)<0$.

Hence, we have a continuous function $f\left(x\right)=\mathrm{cos}\left(x\right)-x$ on the interval [0,1] with $f\left(0\right)=1$ and $f\left(1\right)=\mathrm{cos}\left(1\right)-1<0$. ($a=0$, $b=1$, $f\left(a\right)=1$ and $f\left(b\right)=\mathrm{cos}\left(1\right)-1<0$).

Note that 0 lies in the interval $\left[\mathrm{cos}\left(1\right)-1,1\right]$. Hence, from the intermediate value theorem, there exists a $c\in \left[0,1\right]$ such that f(c)=0.

This means that c is a root of the equation. Hence, we have proved that there exists a root in the interval [0,1].

nidantasnu

You can apply the IVT to the continuous function $x/\mathrm{cos}x$ to show that it takes on the value 1 for some $x$, $0\le x\le 1$.