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

Beginner2022-07-13Added 14 answers

EDIT

Recall the statement of the intermediate value theorem.

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

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

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

We have $f(0)=1>0$.

We also have that $f(1)=\mathrm{cos}(1)-1$. But $\mathrm{cos}(x)<1$, $\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 $\mathrm{cos}(1)<1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(1)<0$.

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

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

Recall the statement of the intermediate value theorem.

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

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

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

We have $f(0)=1>0$.

We also have that $f(1)=\mathrm{cos}(1)-1$. But $\mathrm{cos}(x)<1$, $\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 $\mathrm{cos}(1)<1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}f(1)<0$.

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

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

Beginner2022-07-14Added 7 answers

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

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