Friday, February 10, 2017

Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 49

Use the Intermediate Value Theorem to show that $\cos x = x$ has root on the interval $(0, 1)$

Let $f(x) = x - \cos x$

Based from the definition of Intermediate Value Theorem,
There exist a solution c for the function between the interval $(a,b)$ suppose that the function is continuous on that
given interval. So, there exist a number $c$ between 0 and 1 such that $f(x) = 0$ and that is, $f(c) = 0 $.


$
\begin{equation}
\begin{aligned}

f(0) =& 0 - \cos (0) = -1\\
\\
f(1) =& 1 - \cos (1) = 0.4597


\end{aligned}
\end{equation}
$


By using Intermediate Value Theorem. We prove that...


So,
$
\begin{equation}
\begin{aligned}
& \text{ if } 0 < c < 1 \quad \text{ then } \quad f(0) < f(c) < f(1)\\
& \text{ if } 0 < c < 1 \quad \text{ then } \quad -1 < 0 < 0.4597

\end{aligned}
\end{equation}
$

Therefore,

There exist a such solution $c$ for $x - \cos x = 0 $

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