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$$\Rightarrow f\left(-2\right)=\frac{{\left(-2\right)}^{2}-\left(-2\right)-2}{{\left(-2\right)}^{2}-3\left(-2\right)-8}=\frac{4+2-1}{4+6-8}=\frac{5}{2}>0$$
We summarize steps for drawing sign scheme/ diagram as :
1: Decompose both numerator and denominator into linear factors. Do not cancel common linear factors. Find critical points by equating linear factors individually to zero.
2: Mark distinct critical points on a real number line. If n be the numbers of distinct critical points, then real number line is divided into (n+1) sub-intervals.
3: Test sign of function in a particular interval. Assign alternate signs in adjacent sub-intervals.
4: If a linear factor is repeated even times, then sign of function will not alternate about the critical point corresponding to linear factor in question.
An important point about interpreting sign diagram is that sign of function relates to non-zero values of function. Note that zero does not have sign. The critical points corresponding to numerator function are zeroes of rational function. As such, the graph of function is continuous at these critical points and these critical points can be included in the sub-interval. On the other hand, the rational function is not defined for critical points corresponding to denominator function (as denominator turns zero). We, therefore, conclude that an interval can include critical points corresponding to numerator function, but not the critical points corresponding to denominator function. In case, there are common critical points between numerator and denominator, then those critical points can not be included in the sub-interval.
We can interpret sign diagram in two ways. Either we determine the solution of a given quadratic inequality or we determine intervals of all four types of inequalities for a given quadratic expression. We shall illustrate these two approaches by working with the example case.
Let us consider that we are required to solve rational inequality
$$f\left(x\right)=\frac{{x}^{2}-x-2}{{x}^{2}-3x-8}\ge 0$$
The sign diagram as drawn earlier for the given rational function is shown here :
We need to interpret signs of different intervals to find the solution of a given rational inequality.
Clearly, solution of given inequality is :
$$x\in \left(-\infty ,2\right]U\left(\mathrm{4,}\infty \right)-\{-1,4\}$$
Note that we need to remove -1 and 4 from the solution set as function is not defined for this x – value. However, inequality involved “greater than or equal to” is not strict inequality. It allows equality to zero. As such, we include critical point “2” belonging to numerator function. Further, we can also write the solution set in alternate form as :
$$x\in \left(-\infty ,-1\right)U\left(-\mathrm{1,2}\right]U\left(\mathrm{4,}\infty \right)$$
Let us take the rational expression of example case and determine intervals of each of four inequalities. The sign diagram as drawn earlier is shown here :
$$f\left(x\right)<0;\phantom{\rule{1em}{0ex}}x\in \left(\mathrm{2,4}\right)$$ $$f\left(x\right)\le 0;\phantom{\rule{1em}{0ex}}x\in \left[\mathrm{2,4}\right)$$ $$f\left(x\right)>0;\phantom{\rule{1em}{0ex}}x\in \left(-\infty ,-1\right)U\left(-\mathrm{1,2}\right)U\left(\mathrm{4,}\infty \right)$$ $$f\left(x\right)\ge 0;\phantom{\rule{1em}{0ex}}x\in \left(-\infty ,-1\right)U(-\mathrm{1,2}]U\left(\mathrm{4,}\infty \right)$$
Note that critical point “2” belonging to numerator is included for inequalities which allows equality.
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