Question

Solve each inequality algebraically.

$x^3-x\ge 0$

Short answer

Solution set and the interval is:
$$\begin{gathered} \left\{x\right|-1\le x\le 0orx\ge 1\left\} \\ \right[-1,0]\cup [1,\infty ) \end{gathered}$$

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

$$\begin{gathered} x^3-x=0 \\ x(x^2-1)=0 \\ x=0;(x-1)(x+1)=0 \\ x=0;x=1;x=-1 \end{gathered}$$

Step 2. Form the intervals

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

So the interval we have is:

$(-\infty ,-1)\cup (-1,0)\cup (0,1)\cup (1,\infty )$

Step 3. Form the table

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

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

Interval	Number chosen	Resultant
$(-\infty ,-1)$	-2	$-6\ge 0$
$(-1,0)$	-0.5	role="math" localid="1646200757552" $\frac{3}{8}\ge 0$
$(0,1)$	0.5	role="math" localid="1646200719599" $-\frac{3}{8}\ge 0$
$(1,\infty )$	3	$24\ge 0$

Step 4. Solution set and interval

Since,

$\frac{3}{8}\ge 0,24\ge 0$satisfy $f$. Thus,

$$\begin{gathered} \left\{x\right|-1\le x\le 0orx\ge 1\left\} \\ \right[-1,0]\cup [1,\infty ) \end{gathered}$$