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# Assignment 1 Solutions.pdf

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School
Department
Mathematics
Course
Mathematics 0110A/B
Professor
Chris Brandl
Semester
Fall

Description
MATH 127 Fall 2012 Assignment 1 Solutions 1. Solve the following for x: (a) 3jx + 2j = 2jx ▯ 1j ▯ 1 We have that x+2 = 0 when x = ▯2 and x▯1 = 0 when x = 1. Note that when x = ▯2 the equation becomes 0 = 2j▯3j▯1 = 5 which is not possible. Similarly, when x = 1 the equation becomes 3j3j = 0 ▯ 1 = ▯1 which again is not possible. Thus, x = 1;▯2 are not solutions. We now consider the three intervals x < ▯2, ▯2 < x < 1, and x > 1. For x < ▯2 we have that x + 2 < 0 and x ▯ 1 < 0 so our equation becomes ▯3(x + 2) = ▯2(x ▯ 1) ▯ 1 ▯3x ▯ 6 = ▯2x + 1 ▯3x + 2x = 1 + 6 ▯x = 7 x = ▯7 This value of x lies within our interval and so, x = ▯7 is a solution. Now consider the interval ▯2 < x < 1. For these values of x we get that x+2 > 0 and x ▯ 1 < 0 so our equation becomes 3(x + 2) = ▯2(x ▯ 1) ▯ 1 3x + 6 = ▯2x + 1 3x + 2x = 1 ▯ 6 5x = ▯5 x = ▯1 This value of x lies within our interval and so, x = ▯1 is a solution. 1 Lastly consider the interval x > 1. For these values of x we get that x + 2 > 0 and x ▯ 1 > 0 so our equation becomes 3(x + 2) = 2(x ▯ 1) ▯ 1 3x + 6 = 2x ▯ 3 3x ▯ 2x = ▯3 ▯ 6 x = ▯9 This value of x does not lie in our interval so there are no solutions where x > 1. Hence, the two solutions to the equation are x = ▯7;▯1. (b) jx ▯ 2j = x + 1 We have that x ▯ 2 ▯ 0 for x ▯ 2 and x ▯ 2 < 0 for x < 2. For x ▯ 2 we have x ▯ 2 = x + 1 x ▯ x = 1 + 2 0 = 3 which is impossible and so, there is no solution where x ▯ 2. For x < 2 we have ▯(x ▯ 2) = x + 1 ▯x + 2 = x + 1 ▯x ▯ x = 1 ▯ 2 ▯2x = ▯1 1 x = 2 2. Solve the following inequalities: 2 (a) 4 ▯ 4x ▯ 5 < 7 4 + 5 ▯ 4x < 7 + 5 9 ▯ 4x < 12 9 ▯ x < 3 4 (b) j3 ▯ j < 2 x 4 ▯2 < 3 ▯ x < 2 ▯2 ▯ 3 < ▯ 4 < 2 ▯ 3 x 4 ▯5 < ▯ < ▯1 x 5 1 1 > > 4 x 4 4 < x < 4 5 2 1 (c) > We have that 2x▯3 = 0 when x = 3 and x+6 = 0 when x = ▯6 2x ▯ 3 x + 6 2 so we look at the intervals x < ▯6, ▯6 < x < , and x > . 2 2 Consider when x < ▯6. For these values of x we have that both 2x▯3 and x+6 are negative so when we cross multiply we get that the inequality is ipped twice (stays the same) and so, we have 2(x + 6) > 2x ▯ 3 2x + 12 > 2x ▯ 4 2x ▯ 2x > ▯4 ▯ 12 0 > ▯16 which is always true. Thus, we get that for any x < ▯6 the original inequality holds. 3 Now consider when ▯6 < x < 2. In this interval we get that x+6 is positive and 2x ▯ 3 is negative so when we cross multiply the inequality is ipped and so, we 3 have 2(x + 6) < 2x ▯ 3 2x + 12 < 2x ▯ 4 2x ▯ 2x < ▯4 ▯ 12 0 < ▯16 which is impossible. Thus, there are no solutions in this interval. 3 Lastly consider when x >2. For these values of x we get that both x + 6 and 2x ▯ 3 are positive so when we cross multiply the inequality does not change and so, we have 2(x + 6) > 2x ▯ 3 which we saw gives 0 > ▯16. This is always true so for any x >3the inequality holds. 2 Hence, for any x >3or x < ▯6 the inequality holds. 2 (d) x + 2x ▯ 35 ▯ 0 Factoring the quadratic we get (x + 7)(x ▯ 5) ▯ 0. We see that when x = ▯7 or x = 5 that the inequality is satis▯ed. We now need to consider the three intervals x < ▯7, ▯7 < x < 5, and x > 5. 2 x + 7 x ▯ 5 x
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