Answer
(a) $f(x) = cos~x-x^3$ is continuous for all real numbers.
$f(0) = cos~(0)-(0)^3 = 1 \gt 0$
$f(\pi) = cos~(\pi)-(\pi)^3 = -1-\pi^3 \lt 0$
By the Intermediate Value Theorem, there is a number $c$ in the interval $[0, \pi]$ such that $f(c) = 0$
(b) $f(0.86) = 0.01638$
$f(0.87) = -0.01368$
By the Intermediate Value Theorem, there is a number $c$ in the interval $[0.86, 0.87]$ such that $f(c) = 0$
Work Step by Step
(a) Let $f(x) = cos~x-x^3$
This function is continuous for all real numbers.
$f(0) = cos~(0)-(0)^3 = 1 \gt 0$
$f(\pi) = cos~(\pi)-(\pi)^3 = -1-\pi^3 \lt 0$
By the Intermediate Value Theorem, there is a number $c$ in the interval $[0, \pi]$ such that $f(c) = 0$
(b) We can use trial and error to find a smaller interval that contains a root.
$f(0.80) = cos~(0.80)-(0.80)^3 = 0.1847$
$f(0.90) = cos~(0.90)-(0.90)^3 = -0.10739$
By the Intermediate Value Theorem, there is a number $c$ in the interval $[0.80, 0.90]$ such that $f(c) = 0$
$f(0.86) = cos~(0.86)-(0.86)^3 = 0.01638$
$f(0.87) = cos~(0.87)-(0.87)^3 = -0.01368$
By the Intermediate Value Theorem, there is a number $c$ in the interval $[0.86, 0.87]$ such that $f(c) = 0$
