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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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