Answer
\[f\left( x \right)={{x}^{3}}-x-1\] has a zero between 1 and 2.
Work Step by Step
Find $f\left( 1 \right)$ and $f\left( 2 \right)$. If the sign change in values of $f\left( 1 \right)$ and $f\left( 2 \right)$ takes place, then there must be at least one zero between 1 and 2.
That is,
$\begin{align}
& f\left( x \right)={{x}^{3}}-x-1 \\
& f\left( 1 \right)=1-1-1 \\
& f\left( 1 \right)=-1, \\
\end{align}$
Which is negative.
And
$\begin{align}
& f\left( x \right)={{x}^{3}}-x-1 \\
& f\left( 2 \right)=8-2-1 \\
& f\left( 2 \right)=5, \\
\end{align}$
Which is positive.
Therefore, there is a sign change in the value of $f\left( 1 \right)$ and $f\left( 2 \right)$ and so there is at least one zero between 1 and 2.
