Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Test - Page 137: 13

Answer

$(a)$ $[-4, 3)$ $(b)$ $(-2, 0) ∪ (1, +\infty)$ $(c)$ $(1,7)$ $(d)$ $(-1, 4]$

Work Step by Step

$(a)$ $-4\lt 5-3x\leq 17$ //Subtract $5$ $-9\lt -3x \leq 12$ //Divide by $-3$ $3\gt x \geq -4$ //Arrange the order $-4\leq x \lt 3$ $[-4, 3)$ $(b)$ $x(x-1)(x+2)\gt 0$ We can solve the inequality using intervals. Intervals will be between the points where the expression gets $0$. $x_1=0$ $x_2=1$ $x_3=-2$ So we have four intervals: $(-\infty, -2)$ - Negative $(-2,0)$, - Positive $(0,1)$ - Negative $(1, +\infty)$ - Positive $-2\lt x \lt 0$ and $x \gt 1$ $(-2, 0) ∪ (1, +\infty)$ $(c)$ $|x-4|\lt3$ An inequality involving an absolute value requires two possible solution: $±(x-4)\lt 3$ $(x-4) \lt 3$ $(x-4) \gt -3$ $-3 \lt x-4 \lt 3$ //Add $4$ $1 \lt x \lt 7$ $(1,7)$ $(d)$ $\frac{2x-3}{x+1}\leq1$ //Move all terms to the left-hand side and simplify $\frac{2x-3-x-1}{x+1}\leq 0$ $\frac{x-4}{x+1} \leq 0$ Now we can solve it using intervals method. We have two critical points: $x-4=0$ $x_1=4$ $x+1=0$ $x_2=-1$ //Note, the denominator cannot be negative, so this value corresponds to an open interval So, we have three intervals: $(-\infty, -1)$ - Positive $(-1,4]$ - Negative $[4, +\infty)$ - Positive $-1\lt x \leq 4$ $(-1, 4]$
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