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Stitz-Zeager_College_Algebra_e-book

# At 4 0 no relative or absolute maximum y 4 3 2 1 8 7

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Unformatted text preview: of f , since the y values on the graph extend inﬁnitely upwards. 2. To ﬁnd the zeros of g , we set g (x) = |x − 3| = 0. By Theorem 2.1, we get x − 3 = 0 so that x = 3. Hence, the x-intercept is (3, 0). To ﬁnd our y -intercept, we set x = 0 so that y = g (0) = |0 − 3| = 3, and so (0, 3) is our y -intercept. To graph g (x) = |x − 3|, we use Deﬁnition 2.4 to rewrite as g (x) = |x − 3| = −(x − 3), if x − 3 < 0 x − 3, if x − 3 ≥ 0 Simplifying, we get g (x) = −x + 3, if x < 3 x − 3, if x ≥ 3 130 Linear and Quadratic Functions As before, the open circle we introduce at (3, 0) from the graph of y = −x + 3 is ﬁlled by the point (3, 0) from the line y = x − 3. We determine the domain as (−∞, ∞) and the range as [0, ∞). The function g is increasing on [3, ∞) and decreasing on (−∞, 3]. The relative and absolute minimum value of g is 0 which occurs at (3, 0). As before, there is no relative or absolute maximum value of g . 3. Setting h(x) = 0 to look for zeros gives |x| − 3 = 0. Before we can use any of the properties in Theorem 2.1, we need to isolate the absolute value. Doing so gives |x| = 3 so that x = 3 or x = −3. As a result, we have a pair of x-intercepts: (−3, 0) and (3, 0). Setting x = 0 gives y = h(0) = |0| − 3 = −3, so our y -intercept is (0, −3). As before, we rewrite the absolute value in h to get h(x) = −x − 3, if x < 0 x − 3, if x ≥ 0 Once again, the open circle at (0, −3) from one...
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