Answer
Yes, the function has a real zero between 1 and 2.
Work Step by Step
Substitute $x=1,2$ and evaluate the given function as:
At $x=1$ ,
$\begin{align}
& f\left( 1 \right)={{\left( 1 \right)}^{3}}-1-5 \\
& =1-1-5 \\
& =-5
\end{align}$
At $x=2$ ,
$\begin{align}
& f\left( 2 \right)={{\left( 2 \right)}^{3}}-2-5 \\
& =8-2-5 \\
& =8-7 \\
& =1
\end{align}$
Since between 1 and 2, the sign of the function changes, thus it must cross the x-axis between 1 and 2.
Therefore, for the function $f\left( x \right)={{x}^{3}}-x-5$ , there exists a real zero between 1 and 2.
