Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=x^4-2x^3-2x^2-18x+5$, for $x=3.7$, we have $f(3.7)=(3.7)^4-2(3.7)^3-2(3.7)^2-18(3.7)+5=-2.8999\lt0$
Step 2. For $x=3.8$, we have $f(3.8)=(3.8)^4-2(3.8)^3-2(3.8)^2-18(3.8)+5=6.4896\gt0$
Step 3. As $f(3.7)$ and $f(3.8)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 3.7 and 3.8.