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Stitz-Zeager_College_Algebra_e-book

# Setting the denominator equal to zero gives 2 x 3 1 0

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Unformatted text preview: ution to 2x − 2 = 0. The zeros of r are the solutions to 2x3 − x2 − x = 0, which we have already found to be x = 0, x = − 1 2 and x = 1, the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below. (+) 0 (−) 0 (+) −1 2 0 (+) 1 1 We are interested in where r(x) ≥ 0. We ﬁnd r(x) > 0, or (+), on the intervals −∞, − 2 , (0, 1) and (1, ∞). We add to these intervals the zeros of r, − 1 and 0, to get our ﬁnal solution: 2 −∞, − 1 ∪ [0, 1) ∪ (1, ∞). 2 3 2 1 3. Geometrically, if we set f (x) = x −−x+1 and g (x) = 2 x − 1, the solutions to f (x) = g (x) are x1 the x-coordinates of the points where the graphs of y = f (x) and y = g (x) intersect. The solution to f (x) ≥ g (x) represents not only where the graphs meet, but the intervals over which the graph of y = f (x) is above (>) the graph of g (x). We obtain the graphs below. The ‘Intersect’ command conﬁrms that the graphs cross when x = − 1...
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