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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 $x=1$?

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Can you help me with the steps on using the IVT to solve this? –  soniccool Jun 25 '12 at 1:44
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3 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 $f(x) = \cos(x) - x$.

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 $[0,1]$.

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Can you help me with the steps on using the IVT to solve this? –  soniccool Jun 25 '12 at 1:44
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You can apply the IVT to the continuous function $x/\cos x$ to show that it takes on the value 1 for some $x$, $0\le x\le1$.

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Can you help me with the steps on using the IVT to solve this? –  soniccool Jun 25 '12 at 1:44
    
Maybe we should start with finding out what you know. Can you state the IVT? –  Gerry Myerson Jun 25 '12 at 1:54
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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.

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