Answer
The inequality is valid for values less than -1, values between 0 and 1, and values more than 2 (including them) i.e. $(-\infty,-1]\cap[0,1]\cap [2,\infty)$
Work Step by Step
Now, we find critical points by equating the numerator and denominator to zero:
$x(x^2+1)(x-2)=0$
$(x-1)(x+1)=0$
There are four critical points:
$x_1+1=0\rightarrow x_1=-1$
$(x_2)^2=0\rightarrow x_2=0$
$x_3-1=0\rightarrow x_3=1$
$x_4-2=0\rightarrow x_4=2$
Next, we are going to take five values: one less than -1; one between -1 and 0; one between 0 and 1; one between 1 and 2; and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than -1:
$\dfrac{-2((-2)^2+1)(-2-2)}{(-2-1)(-2+1)}\geq0$
$\dfrac{-2(5)(-4)}{(-3)(-1)}\geq0$
$\dfrac{40}{3}\geq0 \rightarrow \text{ TRUE}$
Second test with a value between -1 and 0:
$\dfrac{-0.5((-0.5)^2+1)(-0.5-2)}{(-0.5-1)(-0.5+1)}\geq0$
$\dfrac{-0.5(1.25)(-2.5)}{(-1.5)(0.5)}\geq0$
$\dfrac{3.125}{-1.5}\geq0 \rightarrow \text{ FALSE}$
Third test with a value between 0 and 1:
$\dfrac{0.5(0.5^2+1)(0.5-2)}{(0.5-1)(0.5+1)}\geq0$
$\dfrac{0.5(1.25)(-1.5)}{(-0.5)(1.5)}\geq0$
$1.25\geq0 \rightarrow \text{ TRUE}$
Fourth test with a value between 1 and 2:
$\dfrac{1.5(1.5^2+1)(1.5-2)}{(1.5-1)(1.5+1)}\geq0$
$\dfrac{1.5(3.25)(-0.5)}{(0.5)(2.5)}\geq0$
$\dfrac{-4.875}{2.5}\geq0 \rightarrow \text{ FALSE}$
Fifth test with a value more than 2:
$\dfrac{3(3^2+1)(3-2)}{(3-1)(3+1)}\geq0$
$\dfrac{3(10)(1)}{(2)(8)}\geq0$
$\dfrac{30}{16}\geq0 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x(x^2+1)(x-2)}{(x-1)(x+1)}\geq0$ is valid for values less than -1, values between 0 and 1, and values more than 2 (including them) i.e. $(-\infty,-1]\cap[0,1]\cap [2,\infty)$
