Question

Solve each inequality algebraically.

$x^3-9x\le 0$

Short answer

Solution set and the interval is:

$$\begin{gathered} \left\{x\right|x\le -3or0\le x\le 3\left\} \\ \right(-\infty ,-3]\cup [0,3) \end{gathered}$$

Step-by-step solution

Step 1. Change the inequality to equal to zero. 

Change the inequality equal to zero to make it easier and then get the value of x

$$\begin{gathered} x^3-9x=0 \\ x(x^2-9)=0 \\ x=0;(x-3)(x+3)=0 \\ x=0;x=3;x=-3 \end{gathered}$$

Step 2. Form the intervals

From the obtained values of x, we can form the interval,

So the interval we have is:

$(-\infty ,-3)\cup (-3,0)\cup (0,3)\cup (3,\infty )$

Step 3. Form the table

Since, $f\left(x\right)\le 0$, so is negative or zero.

check a number in each interval and evaluate the function to see whether it is satisfying the function.

Interval	Number chosen	Resultant
$(-\infty ,-3)$	-4	$-28\le 0$
$(-3,0)$	-1	$8\le 0$
$(0,3)$	2	$-10\le 0$
$(3,\infty )$	4	$28\le 0$

Step 4. Solution set and interval

Since,

$-28\le 0,-10\le 0$satisfy $f$. Thus,

$$\begin{gathered} \left\{x\right|x\le -3or0\le x\le 3\left\} \\ \right(-\infty ,-3]\cup [0,3) \end{gathered}$$