## Precalculus (6th Edition) Blitzer

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.