Intermediate value theorem What steps would I take or use in order to use the intermediate value theorem to show that $\cos x = x$ has a solution between $x=0$ and $x=1$?
 A: 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 $f(x) = \cos(x) - x$.
We have $f(0) = 1 > 0$.
We also have that $f(1) = \cos(1) - 1$. But $\cos(x) < 1$, $\forall x \neq2 n\pi$, where $n \in \mathbb{Z}$. Clearly, $1 \neq 2 n \pi$, where $n \in \mathbb{Z}$. Hence, we have that $\cos(1) < 1 \implies f(1) < 0$.
Hence, we have a continuous function $f(x) = \cos(x) - x$ on the interval $[0,1]$ with $f(0) = 1$ and $f(1) = \cos(1) - 1<0$. ($a=0$, $b=1$, $f(a) = 1$ and $f(b) = \cos(1) -1 < 0$).
Note that $0$ lies in the interval $[\cos(1)-1,1]$. 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 $[0,1]$.
A: You can apply the IVT to the continuous function $x/\cos x$ to show that it takes on the value 1 for some $x$, $0\le x\le1$. 
A: In regards to your recent comments on "using the IVT," I'll elaborate on that.
Others have shown that $f(0) = 1$ and $f(1) \lt 0$. Since there is a sign change (positive to a negative number), then there exists a root within $x =0$ and $x = 1$ by definition of the IVT.
A: In mathematical analysis, the intermediate value theorem states that if a continuous function, $f$, with an interval, $[a, b]$, as its domain, takes values $f(a)$ and $f(b)$ at each end of the interval, then it also takes any value between $f(a)$ and $f(b)$ at some point within the interval.

This has two important corollaries: 


*

*If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem).

*The image of a continuous function over an interval is itself an interval.

Let $f(x)=\cos(x)-x$
$f(0)=1$
$f(1)=\cos(1)-1$
$\therefore f(0)f(1)<0$
Hence, by Bolzano's theorem $f(x)$ must have a root in the interval $(0,1)$.
A: Consider the function $[0,1] \to \mathbb{R}$ given by $f(x)=\cos(x)-x$. Note that $f$ is continuous in $\mathcal{D}(f)=[0,1]$, so you can apply the Intermediate Vale Theorem. Since $f(0)=\cos(0)-0=1>0$ and $f(1)=\cos(1)-1<0$ (since $0<1<\pi/2$, $\cos$ is decreasing in $[0,\pi/2]$ and $\cos(\pi/2)=0$, it follows that $\cos(1)<1$), there exist a number $c \in [0,1]$ such that $$f(c)=0 \Leftrightarrow \cos(c)-c=0 \Leftrightarrow \cos(c)=c.$$ This finishes the proof. 
