## Precalculus (6th Edition) Blitzer

Step 1. Evaluate the given function at $x=1$; we have $f(1)=1^3-2(1)-1=-2$ Step 2. Evaluate the given function at $x=2$; we have $f(2)=2^3-2(2)-1=3$ Step 3. As $f(1)$ and $f(2)$ have opposite signs, based on the Intermediate Value Theorem, there exists a value $1\lt x\lt2$ such that $f(x)=0$