Answer
$[0,1]$.
See graph.
Work Step by Step
Step 1. Let $f(x)\le g(x)$, we have $x^4-1\le x-1 \Longrightarrow x^4-x\le0 \Longrightarrow x(x-1)(x^2+x+1)\le0$, identify boundary points as $x=0,1$.
Step 2. Form intervals $(-\infty,0],[0,1],[1,\infty)$.
Step 3. Choose test values for each 1interval $x=-1,0.5,2$.
Step 4. Test the inequality to get results: $False,\ True,\ False$
Step 5. We have the solution $[0,1]$.
Step 6. See graph.
