I managed to prove this using a direct proof but my prof suggested I try proving it using the contrapositive. Here's what I have so far:
Contrapositive: $(x \le 0) \lor (x \ge 1) \Rightarrow x^4 + 2x^2 - 2x \ge 0$
Splitting this into two, ($P_1 \Rightarrow Q)\land(P_2 \Rightarrow Q)$:
$$x \ge 1 \Rightarrow x^4 + 2x^2 - 2x \ge 0$$
1- $x \ge 1 \Rightarrow [x^4 \ge 1, 2x^2 \ge 2, -2x \ge -2]$
2- $x^4 + 2x^2 - 2x \ge 1 + 2 -2$
3- $x^4 + 2x^2 - 2x \ge 1$
Ok, that was easy enough... but here's where my brain gets stuck.
$$x \le 0 \Rightarrow x^4 + 2x^2 - 2x \ge 0$$
Where should I begin here? In my direct proof I started with the equation and worked towards the x value. I am not sure how to go about proving it the other way. Any help is appreciated.