# 3.5 Rational inequality  (Page 4/6)

 Page 4 / 6

Inclusion and exclusion of critical points

Based on the discussion above, we summarize inclusion or exclusion of critical points here :

1: Question of inclusion of critical points arises when inequality involved is not strict.

2: Critical points belonging to numerator are included in solution set.

3: Critical points belonging to denominator are excluded from solution set.

3: Critical points belonging to both numerator and denominator are excluded from solution set.

## Solution of rational inequalities using wavy curve method

Wavy curve method is a modified sign diagram method. This method has the advantage that we do not need to test sign of interval as required in earlier case. The steps involved are :

1: Factorize numerator and denominator into linear factors.

2: Make coefficients of x positive in all linear factors. This step may require to change sign of “x” in the linear factor by multiplying inequality with -1. Note that this multiplication will change the inequality sign as well. For example, “less than” will become “greater than” etc.

3: Equate each linear factor to zero and find values of x in each case. The values are called critical points.

4: Identify distinct critical points on real number line. The “n” numbers of distinct critical points divide real number lines in (n+1) sub-intervals.

5: The sign of rational function in the right most interval is positive. Alternate sign in adjoining intervals on the left.

5: 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.

We need to exclude exception points i.e. critical points of denominator from solution set. Further, it is important to understand that signs of intervals as determined using this method are not the signs of function – rather signs of modified function in which sign of “x” has changed. However, if we are not required to change the sign of “x” i.e. to modify the function, then signs of intervals are also signs of function. We shall though keep this difference in mind, but we shall refer signs of intervals as sign scheme or diagram in this case also.

Problem : Apply wavy curve method to find the interval of x for the inequality given :

$\frac{x}{1-x}\ge 0$

Solution : We change the sign of "x" in the denominator to positive by multiplying both sides of inequality with -1. Note that this changes the inequality sign as well.

$\frac{x}{x-1}\le 0$

Here, critical points are :

$x=0,1$

The critical points are marked on the real number line. Starting with positive sign in the right most interval, we denote signs of adjacent intervals by alternating sign.

Thus, interval of x as solution of inequality is :

$⇒0\le x<1$

We do not include "1" as it reduces denominator to zero.

Problem : Find solution of the rational inequality given by :

$\frac{3{x}^{2}+6x-15}{\left(2x-1\right)\left(x+3\right)}\ge 1$

Solution : We first convert the given inequality to standard form f(x) ≥ 0.

$⇒\frac{3{x}^{2}+6x-15}{\left(2x-1\right)\left(x+3\right)}-1\ge 0$

$⇒\frac{3{x}^{2}+6x-15-\left(2x-1\right)\left(x+3\right)}{\left(2x-1\right)\left(x+3\right)}\ge 0$

$⇒\frac{3{x}^{2}+6x-15-\left(2{x}^{2}+5x-3\right)}{\left(2x-1\right)\left(x+3\right)}\ge 0$

$⇒\frac{{x}^{2}+x-12}{\left(2x-1\right)\left(x+3\right)}\ge 0$ $⇒\frac{{x}^{2}+4x-3x-12}{\left(2x-1\right)\left(x+3\right)}\ge 0$ $⇒\frac{\left(x-3\right)\left(x+4\right)}{\left(2x-1\right)\left(x+3\right)}\ge 0$

Critical points are -4, -3, 1/2, 3. Corresponding sign diagram is :

The solution of inequality is :

$⇒x\in \left(-\infty ,-4\right]\cup \left(-3,1/2\right)\cup \left[3,\infty \right)$

We do not include "-3" and "1" as they reduce denominator to zero.

#### Questions & Answers

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