Answer
a) $f(-1)=-1 $
$f(0)=3$
There is a root of the equation $f(x) = x^5 − x^2 + 2x+ 3$,
in the interval $(-1, 0)$.
b) There is a root between $-0.88$ and $-0.87$, that is, in the interval
$(0.88, 0.87).$
Work Step by Step
a) Given:
$f(x) = x^5 − x^2 + 2x+ 3$,
The function $f(x) = x^5 − x^2 + 2x+ 3$, which is continuous on the
interval $[-1, 0]$,
$f(-1) = (-1)^5 − (-1)^2 + 2(-1)+ 3$
$=-1−1-2+3 0$
$=-4+3 $
$f(-1)=-1 $
and
$f(0) =(0)^5 − (0)^2 + 2(0)+ 3$
$=0-0+0+3$
$f(0)=3$
Since $3>0 >-1$ there is a number $c$
in $(-1, 0)$ such that $f(c)=0$ by the Intermediate Value Theorem. Thus, there is a root of the equation $ x^5 − x^2 + 2x+ 3=0$,
in the interval $(-1, 0)$.
b)
$f(-0.88) ≈ -0.062 <0$ and $f(-0.87) ≈ 0.0047> 0,$ so there is a root between $-0.88$ and $-0.87$, that is, in the interval
$(0.88, 0.87).$
