#### Answer

See below.

#### Work Step by Step

By plugging in $x=2$ and $x=3$ into $f(x)=x^3-2x-5$, we get $f(2)=2^3-2\cdot2-5=8-4-5=-1$ and $f(3)=3^3-2\cdot3-8=27-6-5=16$
Thus by the Intermediate Value Theorem, the continuous graph must cross the x-axis somewhere between $x=2$ and $x=3$.