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**Unformatted text preview: **Rewriting
the formula for i(x) without absolute values gives i(x) = 4 − 2(−(3x + 1)), if 3x + 1 < 0
=
4 − 2(3x + 1), if 3x + 1 ≥ 0 6x + 6, if x < − 1
3
−6x + 2, if x ≥ − 1
3 The usual analysis near the trouble spot, x = − 1 gives the ‘corner’ of this graph is − 1 , 4 ,
3
3
and we get the distinctive ‘∨’ shape: 2.2 Absolute Value Functions 131
y 3
2
1 −1 1 x −1
−2
−3 i(x) = 4 − 2|3x + 2| The domain of i is (−∞, ∞) while the range is (−∞, 4]. The function i is increasing on
1
1
−∞, − 3 and decreasing on − 3 , ∞ . The relative maximum occurs at the point − 1 , 4
3
and the relative and absolute maximum value of i is 4. Since the graph of i extends downwards
forever more, there is no absolute minimum value. As we can see from the graph, there is no
relative minimum, either.
Note that all of the functions in the previous example bear the characteristic ‘∨’ shape as the graph
of y = |x|. In fact, we could have graphed all o...

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