Answer
The inequality is valid for x-values less than -3 and x-values between 0 and 3 (including them) i.e. $(-\infty,-3]\cap [0,3]$.
Work Step by Step
Now, we find critical points by factoring and equating to zero:
$x^3-9x=0$
$x(x^2-9)=0$
$x(x+3)(x-3)=0$
There are three critical points:
$x_1+3=0\rightarrow x_1=-3$
$x_2=0$
$x_3-3=0\rightarrow x_3=3$
Next, we are going to take four values: one less than -3; one between -3 and 0; one between 0 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than -3:
$(-4)^3-9(-4)\leq0$
$-64-(-36)\leq0$
$-28\leq0 \rightarrow \text{ TRUE}$
Second test with a value between -3 and 0:
$(-1)^3-9(-1)\leq0$
$-1-(-9)\leq0$
$8\leq0 \rightarrow \text{ FALSE}$
Third test with a value between 0 and 3:
$1^3-9(1)\leq0$
$1-9\leq0$
$-8\leq0 \rightarrow \text{ TRUE}$
Fourth test with a value more than 3:
$4^3-9(4)\leq0$
$64-36\leq0$
$28\leq0 \rightarrow \text{ FALSE}$
These tests show that the inequality $x^3-9x\leq0$ is valid for values less than -3 and values between 0 and 3 (including them) i.e. $(-\infty,-3]\cap [0,3]$.
