91Ó°ÊÓ

Given inequality is$\frac{3}{x^2+x-12}>0$

The denominator is to be factored:

$$\begin{gathered} \frac{3}{x^2+x-12}>0 \\ \frac{3}{x^2+4x-3x-12}>0 \\ \frac{3}{x\left(x+4\right)-3\left(x+4\right)}>0 \\ \frac{3}{\left(x+4\right)\left(x-3\right)}>0 \end{gathered}$$

Critical points of a rational expression are the points where the expression becomes 0 or undefined.

The rational expression becomes 0 when the numerator is 0 and undefined when the denominator is 0.

Here the numerator cannot be 0.

When the denominator is 0:

$$\begin{gathered} \left(x+4\right)\left(x-3\right)=0 \\ x+4=0,x-3=0 \\ x=-4,x=3 \end{gathered}$$

Hence the critical points are -4,3

Using the critical points, the number line is divided into intervals. When the inequality is less than or greater than, the boundary values are not included in the interval.

Here the critical points are -4 and 3.

Hence the number line is divided into the intervals$\left(-∞,-4\right),\left(-4,3\right),\left(3,∞\right)$

One value from each interval is to be plugged into the inequality to check which interval satisfies the inequality.

For the interval $\left(-∞,-4\right)$,

Plugging $x=-5$:

$$\begin{gathered} \frac{3}{\left(x+4\right)\left(x-3\right)}>0 \\ \frac{3}{\left(-5+4\right)\left(-5-3\right)}\stackrel{?}{>}0 \\ \frac{3}{\left(-1\right)\left(-8\right)}\stackrel{?}{>}0 \\ \frac{3}{8}>0 \end{gathered}$$

Hence this interval is a solution to the inequality.

For the interval $\left(-4,3\right)$,

Plugging $x=0$:

$$\begin{gathered} \frac{3}{\left(x+4\right)\left(x-3\right)}>0 \\ \frac{3}{\left(0+4\right)\left(0-3\right)}\stackrel{?}{>}0 \\ \frac{3}{\left(4\right)\left(-3\right)}\stackrel{?}{>}0 \\ -\frac{1}{4}≯0 \end{gathered}$$

Hence this interval is not a solution to the inequality.

For the interval $\left(3,∞\right)$,

Plugging $x=4$:

$$\begin{gathered} \frac{3}{\left(x+4\right)\left(x-3\right)}>0 \\ \frac{3}{\left(4+4\right)\left(4-3\right)}\stackrel{?}{>}0 \\ \frac{3}{\left(8\right)\left(1\right)}\stackrel{?}{>}0 \\ \frac{3}{8}>0 \end{gathered}$$

Hence this interval is a solution to the inequality.