Stitz-Zeager_College_Algebra_e-book

X 4 x4 2 x g f x 2x 1 h f x 2x 1 3 i

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Unformatted text preview: (why?), as soon as we found (0, 0) as the x-intercept, we knew this was also the y -intercept. 2.2 Absolute Value Functions 129 y y 4 4 3 3 2 2 1 1 −3 −2 −1 1 2 3 −3 −2 −1 x f (x) = |x|, x < 0 1 2 3 x f (x) = |x|, x ≥ 0 Notice we have an ‘open circle’ at (0, 0) in the graph when x < 0. As we have seen before, this is due to the fact the points on y = −x approach (0, 0) as the x-values approach 0. Since x is required to be strictly less than zero on this stretch, however, the open circle is drawn. However, notice that when x ≥ 0, we get to fill in the point at (0, 0), which effectively ‘plugs’ the hole indicated by the open circle. Hence, we get y 4 3 2 1 −3 −2 −1 1 2 3 x f (x) = |x| By projecting the graph to the x-axis, we see that the domain is (−∞, ∞). Projecting to the y -axis gives us the range [0, ∞). The function is increasing on [0, ∞) and decreasing on (−∞, 0]. The relative minimum value of f is the same as the absolute minimum, namely 0 which occurs at (0, 0). There is no relative maximum value of f . There is also no absolute maximum value...
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