Rational inequalities involve comparing a rational expression to zero or another value. A common method is to move everything to one side, find the zeros of the numerator and the undefined values from the denominator, and use those critical values to divide the number line into intervals. Test the sign on each interval, remembering that zeros may be included for \(\le\) or \(\ge\), while values that make the denominator zero are never included.
Practice Problems
Solve each Rational Inequality
\(\textbf{1)}\) \(\displaystyle \frac{6-x}{x+5}\gt0\)
\(-5 \lt x \lt 6\)
\((-5,6)\)
\(\,\,\,\,\,\,\displaystyle\frac{6-x}{x+5}\gt0\)
\(\,\,\,\,\,\,6-x=0\Rightarrow x=6\)
\(\,\,\,\,\,\,x+5=0\Rightarrow x=-5\)
\(\,\,\,\,\,\,\text{Critical values: }-5,\ 6\)
\(\,\,\,\,\,\,(-\infty,-5):\ –\)
\(\,\,\,\,\,\,(-5,6):\ +\)
\(\,\,\,\,\,\,(6,\infty):\ –\)
\(\,\,\,\,\,\,-5\lt x\lt6\)
The answer is \((-5,6)\)
\(-5 \lt x \lt 6\)
\((-5,6)\)
\(\,\,\,\,\,\,\displaystyle\frac{6-x}{x+5}\gt0\)
\(\,\,\,\,\,\,6-x=0\Rightarrow x=6\)
\(\,\,\,\,\,\,x+5=0\Rightarrow x=-5\)
\(\,\,\,\,\,\,\text{Critical values: }-5,\ 6\)
\(\,\,\,\,\,\,(-\infty,-5):\ –\)
\(\,\,\,\,\,\,(-5,6):\ +\)
\(\,\,\,\,\,\,(6,\infty):\ –\)
\(\,\,\,\,\,\,-5\lt x\lt6\)
The answer is \((-5,6)\)
\(\textbf{2)}\) \(\displaystyle \frac{3x-5}{x-3}\le0\)
\(\frac{5}{3}\le x\lt3\)
\(\left[\frac{5}{3},3\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{3x-5}{x-3}\le0\)
\(\,\,\,\,\,\,3x-5=0\Rightarrow x=\frac{5}{3}\)
\(\,\,\,\,\,\,x-3=0\Rightarrow x=3\)
\(\,\,\,\,\,\,\text{Critical values: }\frac{5}{3},\ 3\)
\(\,\,\,\,\,\,\left(-\infty,\frac{5}{3}\right):\ +\)
\(\,\,\,\,\,\,\left(\frac{5}{3},3\right):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
\(\,\,\,\,\,\,x=\frac{5}{3}\text{ is included}\)
\(\,\,\,\,\,\,x=3\text{ is undefined}\)
The answer is \(\left[\frac{5}{3},3\right)\)
\(\frac{5}{3}\le x\lt3\)
\(\left[\frac{5}{3},3\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{3x-5}{x-3}\le0\)
\(\,\,\,\,\,\,3x-5=0\Rightarrow x=\frac{5}{3}\)
\(\,\,\,\,\,\,x-3=0\Rightarrow x=3\)
\(\,\,\,\,\,\,\text{Critical values: }\frac{5}{3},\ 3\)
\(\,\,\,\,\,\,\left(-\infty,\frac{5}{3}\right):\ +\)
\(\,\,\,\,\,\,\left(\frac{5}{3},3\right):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
\(\,\,\,\,\,\,x=\frac{5}{3}\text{ is included}\)
\(\,\,\,\,\,\,x=3\text{ is undefined}\)
The answer is \(\left[\frac{5}{3},3\right)\)
\(\textbf{3)}\) \(\displaystyle \frac{x^2+3x+2}{x+1}\ge0\)
\(-2\le x\lt-1\text{ or }x\gt-1\)
\([-2,-1)\cup(-1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2+3x+2}{x+1}\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+1)(x+2)}{x+1}\ge0\)
\(\,\,\,\,\,\,x=-2\text{ makes the numerator }0\)
\(\,\,\,\,\,\,x=-1\text{ makes the denominator }0\)
\(\,\,\,\,\,\,(-\infty,-2):\ –\)
\(\,\,\,\,\,\,(-2,-1):\ +\)
\(\,\,\,\,\,\,(-1,\infty):\ +\)
\(\,\,\,\,\,\,x=-2\text{ is included}\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \([-2,-1)\cup(-1,\infty)\)
\(-2\le x\lt-1\text{ or }x\gt-1\)
\([-2,-1)\cup(-1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2+3x+2}{x+1}\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+1)(x+2)}{x+1}\ge0\)
\(\,\,\,\,\,\,x=-2\text{ makes the numerator }0\)
\(\,\,\,\,\,\,x=-1\text{ makes the denominator }0\)
\(\,\,\,\,\,\,(-\infty,-2):\ –\)
\(\,\,\,\,\,\,(-2,-1):\ +\)
\(\,\,\,\,\,\,(-1,\infty):\ +\)
\(\,\,\,\,\,\,x=-2\text{ is included}\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \([-2,-1)\cup(-1,\infty)\)
\(\textbf{4)}\) \(\displaystyle \frac{5x+3}{x+2}\lt-1\)
\(-2\lt x\lt-\frac{5}{6}\)
\(\left(-2,-\frac{5}{6}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3}{x+2}\lt-1\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3}{x+2}+1\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3+x+2}{x+2}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{6x+5}{x+2}\lt0\)
\(\,\,\,\,\,\,6x+5=0\Rightarrow x=-\frac{5}{6}\)
\(\,\,\,\,\,\,x+2=0\Rightarrow x=-2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,\left(-2,-\frac{5}{6}\right):\ –\)
\(\,\,\,\,\,\,\left(-\frac{5}{6},\infty\right):\ +\)
The answer is \(\left(-2,-\frac{5}{6}\right)\)
\(-2\lt x\lt-\frac{5}{6}\)
\(\left(-2,-\frac{5}{6}\right)\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3}{x+2}\lt-1\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3}{x+2}+1\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{5x+3+x+2}{x+2}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{6x+5}{x+2}\lt0\)
\(\,\,\,\,\,\,6x+5=0\Rightarrow x=-\frac{5}{6}\)
\(\,\,\,\,\,\,x+2=0\Rightarrow x=-2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,\left(-2,-\frac{5}{6}\right):\ –\)
\(\,\,\,\,\,\,\left(-\frac{5}{6},\infty\right):\ +\)
The answer is \(\left(-2,-\frac{5}{6}\right)\)
\(\textbf{5)}\) \(\displaystyle \frac{x^3-2x^2}{x+2}\gt0\)
\(x\lt-2\text{ or }x\gt2\)
\((-\infty,-2)\cup(2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^3-2x^2}{x+2}\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2(x-2)}{x+2}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,\ 0,\ 2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,0):\ –\)
\(\,\,\,\,\,\,(0,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=0\text{ and }x=2\text{ give }0\text{ and are excluded}\)
\(\,\,\,\,\,\,x=-2\text{ is undefined}\)
The answer is \((-\infty,-2)\cup(2,\infty)\)
\(x\lt-2\text{ or }x\gt2\)
\((-\infty,-2)\cup(2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x^3-2x^2}{x+2}\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x^2(x-2)}{x+2}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,\ 0,\ 2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,0):\ –\)
\(\,\,\,\,\,\,(0,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=0\text{ and }x=2\text{ give }0\text{ and are excluded}\)
\(\,\,\,\,\,\,x=-2\text{ is undefined}\)
The answer is \((-\infty,-2)\cup(2,\infty)\)
\(\textbf{6)}\) \(\displaystyle\frac{x-4}{x+1}\le0\)
\(-1\lt x\le4\)
\((-1,4]\)
\(\,\,\,\,\,\,\displaystyle\frac{x-4}{x+1}\le0\)
\(\,\,\,\,\,\,x-4=0\Rightarrow x=4\)
\(\,\,\,\,\,\,x+1=0\Rightarrow x=-1\)
\(\,\,\,\,\,\,(-\infty,-1):\ +\)
\(\,\,\,\,\,\,(-1,4):\ –\)
\(\,\,\,\,\,\,(4,\infty):\ +\)
\(\,\,\,\,\,\,x=4\text{ is included}\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \((-1,4]\)
\(-1\lt x\le4\)
\((-1,4]\)
\(\,\,\,\,\,\,\displaystyle\frac{x-4}{x+1}\le0\)
\(\,\,\,\,\,\,x-4=0\Rightarrow x=4\)
\(\,\,\,\,\,\,x+1=0\Rightarrow x=-1\)
\(\,\,\,\,\,\,(-\infty,-1):\ +\)
\(\,\,\,\,\,\,(-1,4):\ –\)
\(\,\,\,\,\,\,(4,\infty):\ +\)
\(\,\,\,\,\,\,x=4\text{ is included}\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \((-1,4]\)
\(\textbf{7)}\) \(\displaystyle\frac{2x+3}{x-5}\gt0\)
\(x\lt-\frac{3}{2}\text{ or }x\gt5\)
\(\left(-\infty,-\frac{3}{2}\right)\cup(5,\infty)\)
\(\,\,\,\,\,\,2x+3=0\Rightarrow x=-\frac{3}{2}\)
\(\,\,\,\,\,\,x-5=0\Rightarrow x=5\)
\(\,\,\,\,\,\,\left(-\infty,-\frac{3}{2}\right):\ +\)
\(\,\,\,\,\,\,\left(-\frac{3}{2},5\right):\ –\)
\(\,\,\,\,\,\,(5,\infty):\ +\)
The answer is \(\left(-\infty,-\frac{3}{2}\right)\cup(5,\infty)\)
\(x\lt-\frac{3}{2}\text{ or }x\gt5\)
\(\left(-\infty,-\frac{3}{2}\right)\cup(5,\infty)\)
\(\,\,\,\,\,\,2x+3=0\Rightarrow x=-\frac{3}{2}\)
\(\,\,\,\,\,\,x-5=0\Rightarrow x=5\)
\(\,\,\,\,\,\,\left(-\infty,-\frac{3}{2}\right):\ +\)
\(\,\,\,\,\,\,\left(-\frac{3}{2},5\right):\ –\)
\(\,\,\,\,\,\,(5,\infty):\ +\)
The answer is \(\left(-\infty,-\frac{3}{2}\right)\cup(5,\infty)\)
\(\textbf{8)}\) \(\displaystyle\frac{(x+2)(x-3)}{x-1}\ge0\)
\(-2\le x\lt1\text{ or }x\ge3\)
\([-2,1)\cup[3,\infty)\)
\(\,\,\,\,\,\,\text{Critical values: }-2,\ 1,\ 3\)
\(\,\,\,\,\,\,(-\infty,-2):\ –\)
\(\,\,\,\,\,\,(-2,1):\ +\)
\(\,\,\,\,\,\,(1,3):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
\(\,\,\,\,\,\,x=-2\text{ and }x=3\text{ are included}\)
\(\,\,\,\,\,\,x=1\text{ is excluded}\)
The answer is \([-2,1)\cup[3,\infty)\)
\(-2\le x\lt1\text{ or }x\ge3\)
\([-2,1)\cup[3,\infty)\)
\(\,\,\,\,\,\,\text{Critical values: }-2,\ 1,\ 3\)
\(\,\,\,\,\,\,(-\infty,-2):\ –\)
\(\,\,\,\,\,\,(-2,1):\ +\)
\(\,\,\,\,\,\,(1,3):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
\(\,\,\,\,\,\,x=-2\text{ and }x=3\text{ are included}\)
\(\,\,\,\,\,\,x=1\text{ is excluded}\)
The answer is \([-2,1)\cup[3,\infty)\)
\(\textbf{9)}\) \(\displaystyle\frac{x-1}{x^2-9}\lt0\)
\(x\lt-3\text{ or }1\lt x\lt3\)
\((-\infty,-3)\cup(1,3)\)
\(\,\,\,\,\,\,\displaystyle\frac{x-1}{(x-3)(x+3)}\lt0\)
\(\,\,\,\,\,\,\text{Critical values: }-3,\ 1,\ 3\)
\(\,\,\,\,\,\,(-\infty,-3):\ –\)
\(\,\,\,\,\,\,(-3,1):\ +\)
\(\,\,\,\,\,\,(1,3):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
The answer is \((-\infty,-3)\cup(1,3)\)
\(x\lt-3\text{ or }1\lt x\lt3\)
\((-\infty,-3)\cup(1,3)\)
\(\,\,\,\,\,\,\displaystyle\frac{x-1}{(x-3)(x+3)}\lt0\)
\(\,\,\,\,\,\,\text{Critical values: }-3,\ 1,\ 3\)
\(\,\,\,\,\,\,(-\infty,-3):\ –\)
\(\,\,\,\,\,\,(-3,1):\ +\)
\(\,\,\,\,\,\,(1,3):\ –\)
\(\,\,\,\,\,\,(3,\infty):\ +\)
The answer is \((-\infty,-3)\cup(1,3)\)
\(\textbf{10)}\) \(\displaystyle\frac{x+4}{x-2}\ge1\)
\(x\gt2\)
\((2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4}{x-2}\ge1\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4}{x-2}-1\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4-(x-2)}{x-2}\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{6}{x-2}\ge0\)
\(\,\,\,\,\,\,x=2\text{ is the only critical value}\)
\(\,\,\,\,\,\,(-\infty,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
The answer is \((2,\infty)\)
\(x\gt2\)
\((2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4}{x-2}\ge1\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4}{x-2}-1\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+4-(x-2)}{x-2}\ge0\)
\(\,\,\,\,\,\,\displaystyle\frac{6}{x-2}\ge0\)
\(\,\,\,\,\,\,x=2\text{ is the only critical value}\)
\(\,\,\,\,\,\,(-\infty,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
The answer is \((2,\infty)\)
\(\textbf{11)}\) \(\displaystyle\frac{2x-1}{x+3}\lt2\)
\(x\gt-3\)
\((-3,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{2x-1}{x+3}\lt2\)
\(\,\,\,\,\,\,\displaystyle\frac{2x-1-2(x+3)}{x+3}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{-7}{x+3}\lt0\)
\(\,\,\,\,\,\,x=-3\text{ is the critical value}\)
\(\,\,\,\,\,\,(-\infty,-3):\ +\)
\(\,\,\,\,\,\,(-3,\infty):\ –\)
The answer is \((-3,\infty)\)
\(x\gt-3\)
\((-3,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{2x-1}{x+3}\lt2\)
\(\,\,\,\,\,\,\displaystyle\frac{2x-1-2(x+3)}{x+3}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{-7}{x+3}\lt0\)
\(\,\,\,\,\,\,x=-3\text{ is the critical value}\)
\(\,\,\,\,\,\,(-\infty,-3):\ +\)
\(\,\,\,\,\,\,(-3,\infty):\ –\)
The answer is \((-3,\infty)\)
\(\textbf{12)}\) \(\displaystyle\frac{x^2-4}{x^2-1}\gt0\)
\(x\lt-2\text{ or }-1\lt x\lt1\text{ or }x\gt2\)
\((-\infty,-2)\cup(-1,1)\cup(2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)(x+2)}{(x-1)(x+1)}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,-1,1,2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,-1):\ –\)
\(\,\,\,\,\,\,(-1,1):\ +\)
\(\,\,\,\,\,\,(1,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
The answer is \((-\infty,-2)\cup(-1,1)\cup(2,\infty)\)
\(x\lt-2\text{ or }-1\lt x\lt1\text{ or }x\gt2\)
\((-\infty,-2)\cup(-1,1)\cup(2,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)(x+2)}{(x-1)(x+1)}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,-1,1,2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,-1):\ –\)
\(\,\,\,\,\,\,(-1,1):\ +\)
\(\,\,\,\,\,\,(1,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
The answer is \((-\infty,-2)\cup(-1,1)\cup(2,\infty)\)
\(\textbf{13)}\) \(\displaystyle\frac{x^2-9}{x+3}\le0\)
\(x\le3,\quad x\ne-3\)
\((-\infty,-3)\cup(-3,3]\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-3)(x+3)}{x+3}\le0\)
\(\,\,\,\,\,\,x-3\le0,\quad x\ne-3\)
\(\,\,\,\,\,\,x\le3,\quad x\ne-3\)
\(\,\,\,\,\,\,x=-3\text{ is a hole and must be excluded}\)
The answer is \((-\infty,-3)\cup(-3,3]\)
\(x\le3,\quad x\ne-3\)
\((-\infty,-3)\cup(-3,3]\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-3)(x+3)}{x+3}\le0\)
\(\,\,\,\,\,\,x-3\le0,\quad x\ne-3\)
\(\,\,\,\,\,\,x\le3,\quad x\ne-3\)
\(\,\,\,\,\,\,x=-3\text{ is a hole and must be excluded}\)
The answer is \((-\infty,-3)\cup(-3,3]\)
\(\textbf{14)}\) \(\displaystyle\frac{(x-2)^2}{x+1}\ge0\)
\(x\gt-1\)
\((-1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)^2}{x+1}\ge0\)
\(\,\,\,\,\,\,\text{Critical values: }-1,\ 2\)
\(\,\,\,\,\,\,(x-2)^2\ge0\text{ for all }x\)
\(\,\,\,\,\,\,(-\infty,-1):\ –\)
\(\,\,\,\,\,\,(-1,2):\ +\)
\(\,\,\,\,\,\,x=2:\ 0\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \((-1,\infty)\)
\(x\gt-1\)
\((-1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)^2}{x+1}\ge0\)
\(\,\,\,\,\,\,\text{Critical values: }-1,\ 2\)
\(\,\,\,\,\,\,(x-2)^2\ge0\text{ for all }x\)
\(\,\,\,\,\,\,(-\infty,-1):\ –\)
\(\,\,\,\,\,\,(-1,2):\ +\)
\(\,\,\,\,\,\,x=2:\ 0\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=-1\text{ is excluded}\)
The answer is \((-1,\infty)\)
\(\textbf{15)}\) \(\displaystyle\frac{x+1}{(x-2)^2}\le0\)
\(x\le-1\)
\((-\infty,-1]\)
\(\,\,\,\,\,\,\displaystyle\frac{x+1}{(x-2)^2}\le0\)
\(\,\,\,\,\,\,(x-2)^2\gt0\text{ for }x\ne2\)
\(\,\,\,\,\,\,x+1\le0\)
\(\,\,\,\,\,\,x\le-1\)
\(\,\,\,\,\,\,x=2\text{ is undefined but is not in the solution interval}\)
The answer is \((-\infty,-1]\)
\(x\le-1\)
\((-\infty,-1]\)
\(\,\,\,\,\,\,\displaystyle\frac{x+1}{(x-2)^2}\le0\)
\(\,\,\,\,\,\,(x-2)^2\gt0\text{ for }x\ne2\)
\(\,\,\,\,\,\,x+1\le0\)
\(\,\,\,\,\,\,x\le-1\)
\(\,\,\,\,\,\,x=2\text{ is undefined but is not in the solution interval}\)
The answer is \((-\infty,-1]\)
\(\textbf{16)}\) \(\displaystyle\frac{x^2+x-6}{x^2-4}\lt0\)
\(-3\lt x\lt-2\)
\((-3,-2)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+3)(x-2)}{(x-2)(x+2)}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+3}{x+2}\lt0,\quad x\ne2\)
\(\,\,\,\,\,\,\text{Critical values: }-3,\ -2\)
\(\,\,\,\,\,\,(-\infty,-3):\ +\)
\(\,\,\,\,\,\,(-3,-2):\ –\)
\(\,\,\,\,\,\,(-2,\infty):\ +\)
The answer is \((-3,-2)\)
\(-3\lt x\lt-2\)
\((-3,-2)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x+3)(x-2)}{(x-2)(x+2)}\lt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+3}{x+2}\lt0,\quad x\ne2\)
\(\,\,\,\,\,\,\text{Critical values: }-3,\ -2\)
\(\,\,\,\,\,\,(-\infty,-3):\ +\)
\(\,\,\,\,\,\,(-3,-2):\ –\)
\(\,\,\,\,\,\,(-2,\infty):\ +\)
The answer is \((-3,-2)\)
\(\textbf{17)}\) \(\displaystyle\frac{x^2-5x+6}{x-4}\ge0\)
\(2\le x\le3\text{ or }x\gt4\)
\([2,3]\cup(4,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)(x-3)}{x-4}\ge0\)
\(\,\,\,\,\,\,\text{Critical values: }2,\ 3,\ 4\)
\(\,\,\,\,\,\,(-\infty,2):\ –\)
\(\,\,\,\,\,\,(2,3):\ +\)
\(\,\,\,\,\,\,(3,4):\ –\)
\(\,\,\,\,\,\,(4,\infty):\ +\)
\(\,\,\,\,\,\,x=2\text{ and }x=3\text{ are included}\)
\(\,\,\,\,\,\,x=4\text{ is excluded}\)
The answer is \([2,3]\cup(4,\infty)\)
\(2\le x\le3\text{ or }x\gt4\)
\([2,3]\cup(4,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-2)(x-3)}{x-4}\ge0\)
\(\,\,\,\,\,\,\text{Critical values: }2,\ 3,\ 4\)
\(\,\,\,\,\,\,(-\infty,2):\ –\)
\(\,\,\,\,\,\,(2,3):\ +\)
\(\,\,\,\,\,\,(3,4):\ –\)
\(\,\,\,\,\,\,(4,\infty):\ +\)
\(\,\,\,\,\,\,x=2\text{ and }x=3\text{ are included}\)
\(\,\,\,\,\,\,x=4\text{ is excluded}\)
The answer is \([2,3]\cup(4,\infty)\)
\(\textbf{18)}\) \(\displaystyle\frac{x-5}{x^2-4x+3}\gt0\)
\(1\lt x\lt3\text{ or }x\gt5\)
\((1,3)\cup(5,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x-5}{(x-1)(x-3)}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }1,\ 3,\ 5\)
\(\,\,\,\,\,\,(-\infty,1):\ –\)
\(\,\,\,\,\,\,(1,3):\ +\)
\(\,\,\,\,\,\,(3,5):\ –\)
\(\,\,\,\,\,\,(5,\infty):\ +\)
The answer is \((1,3)\cup(5,\infty)\)
\(1\lt x\lt3\text{ or }x\gt5\)
\((1,3)\cup(5,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x-5}{(x-1)(x-3)}\gt0\)
\(\,\,\,\,\,\,\text{Critical values: }1,\ 3,\ 5\)
\(\,\,\,\,\,\,(-\infty,1):\ –\)
\(\,\,\,\,\,\,(1,3):\ +\)
\(\,\,\,\,\,\,(3,5):\ –\)
\(\,\,\,\,\,\,(5,\infty):\ +\)
The answer is \((1,3)\cup(5,\infty)\)
\(\textbf{19)}\) \(\displaystyle\frac{x^2-1}{x^2-4}\le0\)
\(-2\lt x\le-1\text{ or }1\le x\lt2\)
\((-2,-1]\cup[1,2)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-1)(x+1)}{(x-2)(x+2)}\le0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,-1,1,2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,-1):\ –\)
\(\,\,\,\,\,\,(-1,1):\ +\)
\(\,\,\,\,\,\,(1,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=-1\text{ and }x=1\text{ are included}\)
\(\,\,\,\,\,\,x=-2\text{ and }x=2\text{ are excluded}\)
The answer is \((-2,-1]\cup[1,2)\)
\(-2\lt x\le-1\text{ or }1\le x\lt2\)
\((-2,-1]\cup[1,2)\)
\(\,\,\,\,\,\,\displaystyle\frac{(x-1)(x+1)}{(x-2)(x+2)}\le0\)
\(\,\,\,\,\,\,\text{Critical values: }-2,-1,1,2\)
\(\,\,\,\,\,\,(-\infty,-2):\ +\)
\(\,\,\,\,\,\,(-2,-1):\ –\)
\(\,\,\,\,\,\,(-1,1):\ +\)
\(\,\,\,\,\,\,(1,2):\ –\)
\(\,\,\,\,\,\,(2,\infty):\ +\)
\(\,\,\,\,\,\,x=-1\text{ and }x=1\text{ are included}\)
\(\,\,\,\,\,\,x=-2\text{ and }x=2\text{ are excluded}\)
The answer is \((-2,-1]\cup[1,2)\)
\(\textbf{20)}\) \(\displaystyle\frac{x+2}{x-1}+1\gt0\)
\(x\lt-\frac{1}{2}\text{ or }x\gt1\)
\(\left(-\infty,-\frac{1}{2}\right)\cup(1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x+2}{x-1}+1\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+2+x-1}{x-1}\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{2x+1}{x-1}\gt0\)
\(\,\,\,\,\,\,2x+1=0\Rightarrow x=-\frac{1}{2}\)
\(\,\,\,\,\,\,x-1=0\Rightarrow x=1\)
\(\,\,\,\,\,\,\left(-\infty,-\frac{1}{2}\right):\ +\)
\(\,\,\,\,\,\,\left(-\frac{1}{2},1\right):\ –\)
\(\,\,\,\,\,\,(1,\infty):\ +\)
The answer is \(\left(-\infty,-\frac{1}{2}\right)\cup(1,\infty)\)
\(x\lt-\frac{1}{2}\text{ or }x\gt1\)
\(\left(-\infty,-\frac{1}{2}\right)\cup(1,\infty)\)
\(\,\,\,\,\,\,\displaystyle\frac{x+2}{x-1}+1\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{x+2+x-1}{x-1}\gt0\)
\(\,\,\,\,\,\,\displaystyle\frac{2x+1}{x-1}\gt0\)
\(\,\,\,\,\,\,2x+1=0\Rightarrow x=-\frac{1}{2}\)
\(\,\,\,\,\,\,x-1=0\Rightarrow x=1\)
\(\,\,\,\,\,\,\left(-\infty,-\frac{1}{2}\right):\ +\)
\(\,\,\,\,\,\,\left(-\frac{1}{2},1\right):\ –\)
\(\,\,\,\,\,\,(1,\infty):\ +\)
The answer is \(\left(-\infty,-\frac{1}{2}\right)\cup(1,\infty)\)
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\(\bullet\text{ Partial Fraction Decomposition}\)
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