Answer
There is a zero in the interval $(2.75,3)$
Work Step by Step
We are given the function:
$f(x)=x^3-8x-1$
Compute the value of $f$ for $x=2$ and $x=3$:
$f(2)=2^3-8(2)-1=-9$
$f(3)=3^3-8(3)-1=2$
As $f(2)f(3)=-9(2)=-18<0$, $f$ has a zero in the interval $(2,3)$.
The midpoint of the interval $(2,3)$ is $2.5$. Compute $f$ in $x=2.5$:
$f(2.5)=2.5^3-8(2.5)-1=-5.375$
As $f(2.5)f(3)<0$, there is a zero in the interval $(2.5,3)$.
The midpoint of the interval $(2.5,3)$ is $2.75$. Compute $f$ in $x=2.75$:
$f(2.75)=2.75^3-8(2.75)-1=-2.203125$
As $f(2.75)f(3)<0$, there is a zero in the interval $(2.75,3)$.
