Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=2x^4-4x^2+4x-8$, for $x=1$, we have $f(1)=2(1)^4-4(1)^2+4(1)-8=-6\lt0$
Step 2. For $x=2$, we have $f(2)=2(2)^4-4(2)^2+4(2)-8=16\gt0$
Step 3. As $f(1)$ and $f(2)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 1 and 2.