Single Variable Calculus, Chapter 2, 2.5, Section 2.5, Problem 51

(a) Show that the equation $\cos x = x^3$ has at least one real root.
(b) Determine an inteval of length 0.01 that contains a root using a calculator.

(a) Let $f(x) = x^3 - \cos x$
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 the given interval. So we take $a$ and $b$ to be -1 and 1 respectively and assume the function $f(x)$
is continuous on the interval (-1,1). So we have,


$
\begin{equation}
\begin{aligned}

f(-1) =& x^3 - \cos x = (-1)^3 - \cos (-1) = -1.54\\
\\
f(1) =& x^3 - \cos x = (1)^3 - \cos (1) = 0.46

\end{aligned}
\end{equation}
$


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

$
\begin{equation}
\begin{aligned}
& \text{if } -1 < c < 1 && \text{then } \quad f(-1) < f(c) < f(1)\\
& \text{So, }\\
& \text{if } -1 < c < 1 && \text{then } \quad -1.54 < 0 < 0.46
\end{aligned}
\end{equation}
$

Therefore,
There exist such root for $x^3 - \cos x = 0$.


(b) By trial and error using calculator, we take the interval (0.86,0.87) so,

$
\begin{equation}
\begin{aligned}

f(0.86) =& (0.86)^3 - \cos (0.86) = -0.016 < 0\\
& \text{and}\\
f(0.87) =& (0.87)^3 - \cos (0.87) = 0.014 > 0

\end{aligned}
\end{equation}
$


Therefore,
The root in the function $x^3 - \cos x = 0$ exists between (0.86,0.87).