## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $\frac{(2+x)(2-x)}{x(x-1)} \geq0$ INTERVALS $[-2,0) (1,2]$ GRAPH Put four numbers in the real number line -2, 0, 1 and 2. 0 and 1 should be with open circle. -2 and 2 with close circlel. Connect -2 and 0. Also, connect 1 and 2.
$\frac{3}{x-1}-\frac{4}{x}\geq1$ $\frac{3}{x-1}-\frac{4}{x}-1\geq0$ $\frac{3x-4(x-1)-1(x^{2}-x)}{x(x-1)}\geq0$ $\frac{3x-4x+4-x^{2}+x}{x(x-1)} \geq0$ $\frac{4-x^{2}}{x(x-1)}\geq0$ $\frac{(2+x)(2-x)}{x(x-1)} \geq0$ Key numbers NUMERATOR: x=2 x=-2 DENOMINATOR x=0 x=1 INTERVALS $(-\infty,-2] [-2,0) (0,1) (1,2] [2,\infty)$ test the values $(-\infty,-2]= y(-3)=\frac{(2-3)(2+3)}{-3(-3-1)} \geq0= y(-3)=-0.41$ no solution [-2,0)= $y(-1)=\frac{(2-1)(2+1)}{-1(-1-1)} \geq0= y(-1)=1.5$ Solution (0,1)= y(0.5)=\frac{(2+0.5)(2-0.5)}{0.5(0.5-1)} \geq0= y(0.5)=-15$No solution (1,2]= y(1.5)=\frac{2+1.5)(2-1.5)}{1.5(1.5-1)} \geq0= y(1.5)=2.3$ Solution $[2,\infty)$= $y(4)=\frac{(2+4)(2-4)}{2(2-1)} \geq0= Y(4)=-8$ No solution SOLUTION $[-2,0) (1,2]$ GRAPH Put four numbers in the real number line -2, 0, 1 and 2. 0 and 1 should be with open circle. -2 and 2 with close circlel. Connect -2 and 0. Also, connect 1 and 2.