Answer
$f(0)=1 >0$
$f(1)\approx-0.46<0$
There is a root of the equation $cos x − x^3 = 0.$ or
$cos x = x^3$. in the interval $(0, 1)$.
b) $f(0.86) ≈ 0.016 > 0$ and $f(0.87) ≈ −0.014 < 0,$ so there is a root between $0.86$ and $0.87$, that is, in the interval
$(0.86, 0.87).$
Work Step by Step
a) Given:
$f(x) = cos x − x^3$,
$f(x) = cos x − x^3$, which is continous on the
interval $[0, 1]$,
$f(0) =cos~ 0− 0^3 $
$=1− 0$
$f(0)=1 >0$
and
$f(1) =cos~1 -1^3$
$f(1)\approx-0.46<0$
Since $1>0 >-0.46$ there is a number $c$
in $(1, 2)$ such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a root of the equation $cos x − x^3 = 0.$ or
$cos x = x^3$. in the interval $(0, 1)$.
b)
$f(0.86) ≈ 0.016 > 0$ and $f(0.87) ≈ −0.014 < 0,$ so there is a root between $0.86$ and $0.87$, that is, in the interval
$(0.86, 0.87).$
