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Stitz-Zeager_College_Algebra_e-book

1 solution 1 the formula f x x2 4x 3 is in the form

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Unformatted text preview: using cases by re-writing the function g with two separate applications of Deﬁnition 2.4 to remove each instance of the absolute values, one at a time. In the ﬁrst round we get g (x) = 2 −(x + 2) − |x − 3| + 1, if x + 2 < 0 = (x + 2) − |x − 3| + 1, if x + 2 ≥ 0 −x − 1 − |x − 3|, if x < −2 x + 3 − |x − 3|, if x ≥ −2 These are set braces, not parentheses or brackets. We used this same ‘set builder’ notation in Section 1.4. 2.2 Absolute Value Functions 133 Given that |x − 3| = −(x − 3), if x − 3 < 0 = x − 3, if x − 3 ≥ 0 −x + 3, if x < 3 , x − 3, if x ≥ 3 we need to break up the domain again at x = 3. Note that if x < −2, then x < 3, so we replace |x − 3| with −x + 3 for that part of the domain, too. Our completed revision of the form of g yields −x − 1 − (−x + 3), if x < −2 g (x) = x + 3 − (−x + 3), if x ≥ −2 and x < 3 x + 3 − (x − 3), if x ≥ 3 −4, if x < −2 = 2x, if −2 ≤ x < 3 6, if x ≥ 3 To solve g (x) = 0, we see...
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