Stitz-Zeager_College_Algebra_e-book

2 d a 3 b 4 c 153 90 e 120 b 3 c

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Unformatted text preview: x) and cos(x) are all real numbers, the only concern when finding sin(x the domain of f (x) = 2 cos(x))−1 is division by zero so we set the denominator equal to zero and π solve. From 2 cos(x) − 1 = 0 we get cos(x) = 1 so that x = π +2πk or x = 53 +2πk for integers 2 3 π k . Using set-builder notation, the domain is x : x = π + 2πk and x = 53 + 2πk for integers k . 3 π 5π 7π 11π Writing out a few of the terms gives x : x = ± 3 , ± 3 , ± 3 , ± 3 , . . . , so we have − 11π 3 π − 73 π − 53 −π 3 π 3 5π 3 7π 3 11π 3 Unlike the previous example, we have two different families of points to consider, and we present two ways of dealing with this kind of situation. One way is to generalize what we did in the previous example and use the formulas we found in our domain work to describe π the intervals. To that end, we let ak = π + 2πk = (6k+1)π and bk = 53 + 2πk = (6k+5)π for 3 3 3 integers k . The goal now is to write the domain in terms of the a’s an b’s. We find a0 = π , 3 π π π a1 = 73 , a−1 = − 53 , a2 = 13π , a−2 = − 11π , b0 = 53...
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