Answer
The inequality is valid for x-values between -1 and 0 and x-values more than 1 (including them) i.e. $[-1,0]\cap [1,\infty)$.
Work Step by Step
Now, we find critical points by factoring and equating to zero:
$x^3-x=0$
$x(x^2-1)=0$
$x(x+1)(x-1)=0$
There are three critical points:
$x_1+1=0\rightarrow x_1=-1$
$x_2=0$
$x_3-1=0\rightarrow x_3=1$
Next, we are going to take four values: one less than -1; one between -1 and 0; one between 0 and 1; and one more than 1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$(-2)^3-(-2)\geq0$
$-8+2\geq0$
$-6\geq0 \rightarrow \text{ FALSE}$
Second test with a value between -1 and 0:
$(-0.2)^3-(-0.2)\geq0$
$-0.008+0.2\geq0$
$0.192\geq0 \rightarrow \text{ TRUE}$
Third test with a value between 0 and 1:
$0.2^3-0.2\geq0$
$0.008-0.2\geq0$
$-0.192\geq0 \rightarrow \text{ FALSE}$
Fourth test with a value more than 1:
$2^3-2\geq0$
$8-2\geq0$
$6\geq0 \rightarrow \text{ TRUE}$
These tests show that the inequality $x^3-x\geq0$ is valid for values between -1 and 0 and value more than 1 (including them) i.e. $[-1,0]\cap [1,\infty)$
