Stitz-Zeager_College_Algebra_e-book

Every point x 1 where x 4 relative maximum and minimum

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Unformatted text preview: piece of the graph of h is filled by the point (0, −3) from the other piece of h. From the graph, we determine the domain of h is (−∞, ∞) and the range is [−3, ∞). On [0, ∞), h is increasing; on (−∞, 0] it is decreasing. The relative minimum occurs at the point (0, −3) on the graph, and we see −3 is both the relative and absolute minimum value of h. Also, h has no relative or absolute maximum value. y y 1 4 3 −3 −2 −1 1 2 3 x −1 2 −2 1 −3 1 2 3 4 5 x g (x) = |x − 3| −4 h(x) = |x| − 3 4. As before, we set i(x) = 0 to find the zeros of i to get 4 − 2|3x + 1| = 0. Once again, we need to isolate the absolute value expression before we apply Theorem 2.1. So 4 − 2|3x + 1| = 0 becomes 2|3x + 1| = 4 and hence, |3x + 1| = 2. Applying Theorem 2.1, we get 3x + 1 = 2 or 1 3x + 1 = −2, from which we get x = 3 or x = −1. Our x-intercepts are 1 , 0 and (−1, 0). 3 Substituting x = 0 gives y = i(0) = 4 − 2|3(0) + 1| = 2, for a y -intercept of (0, 2)....
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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