Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.8 - Inequalities - 1.8 Exercises: 71

Answer

SOLUTION $\frac{-2x-1}{(x+3)(x-2)} \lt0$ INTERVAL $(-3,\frac{-1}{2}] (2,\infty)$ GRAPH In the number line put 3 numbers -3, $\frac{-1}{2}$ and 2. You shoould put a close circle $\frac{-1}{2}$ and an opne circle in -3 and 2. You should connect -3 and $\frac{-1}{2}$. Also draw a line on 2 and pointed to $\infty$.

Work Step by Step

$\frac{x+2}{x+3} \lt\frac{x-1}{x-2}$ $\frac{x+2}{x+3}-\frac{x-1}{x-2} \lt0$ $\frac{(x+2)(x-2)-(x-1)(x+3)}{(x+3)(x-2)} \lt0$ $\frac{x^{2}-4-(x^{2}+2x-3)}{(x+3)(x-2)} \lt0$ $\frac{x^{2}-4-x^{2}-2x+3}{(x+3)(x-2)} \lt0$ $\frac{-2x-1}{(x+3)(x-2)} \lt0$ Now you have to look for the solutions given by the inequality. NUMERATOR $x=\frac{-1}{2}$ DENOMINATOR x=-3 x=2 INTERVALS Options: $(\infty,-3) (-3,\frac{-1}{2}] [\frac{-1}{2},2) (2,\infty)$ $(\infty,-3)$ Y(-4)=\frac{-2(-4)-1}{(-4+3)(-4-2)} \lt0= y(-4)=1.16 No solution $(-3,\frac{-1}{2}]$ $Y(-2)=\frac{-2(-2)-1}{(-2+3)(-2-2)} \lt0= y(-2)=-0.75$ Solution $[\frac{-1}{2},2)$ $y(0)=\frac{-2(0)-1}{(0+3)(0-2)} \lt0= y(0)=0.16$ No solution $(2,\infty)$ $Y(5)=\frac{-2(5)-1}{(5+3)(5-2)} \lt0= y(5)=-0.458$ Solution CORRECT $(-3,\frac{-1}{2}] (2,\infty)$ GRAPH In the number line put 3 numbers -3, $\frac{-1}{2}$ and 2. You shoould put a close circle $\frac{-1}{2}$ and an opne circle in -3 and 2. You should connect -3 and $\frac{-1}{2}$. Also draw a line on 2 and pointed to $\infty$.
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