Consider the following.

cos$($x$)=$ x$3$

$($a$)$ Prove that the equation has at least one real root.

The equation

cos$($x$)=$ x$3$

is equivalent to the equation

f$($x$)=$ cos$($x$)$ x$3=0.$

f$($x$)$

is continuous on the interval

$[0,1],$

f$(0)=$

$,$

and

f$(1)=$

$.$

Since

$---$Select$---$ f$(0)$ f$(1)<0<---$Select$---$ f$(0)$ f$(1),$

there is a number c in

$(0,1)$

such that

f$($c$)=0$

by the Intermediate Value Theorem. Thus, there is a root of the equation

cos$($x$)=$ x$3,$

in the interval

$(0,1).$

$($b$)$ Use your calculator to find an interval of length $0\mathrm{.}01$ that contains a root. $($Enter your answer using interval notation. Round your answers to two decimal places.$)$