Stitz-Zeager_College_Algebra_e-book

287x 2473908 solving for x yields x 2015454 since the

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Unformatted text preview: find the only one zero of f , x = 2 . This one x value divides the number line into two 3 intervals, from which we choose x = 0 and x = 1 as test values. We find f (0) = 4 > 0 and f (1) = 1 > 0. Since we are looking for solutions to 9x2 − 12x + 4 ≤ 0, we are looking for 2.4 Inequalities 163 x values where 9x2 − 12x + 4 < 0 as well as where 9x2 − 12x + 4 = 0. Looking at our sign diagram, there are no places where 9x2 − 12x + 4 < 0 (there are no (−)), so our solution 2 is only x = 3 (where 9x2 − 12x + 4 = 0). We write this as 2 . Graphically, we solve 3 9x2 + 4 ≤ 12x by graphing g (x) = 9x2 + 4 and h(x) = 12x. We are looking for the x values where the graph of g is below the graph of h (for 9x2 + 4 < 12x) and where the two graphs 2 intersect (9x2 + 4 = 12x). We see the line and the parabola touch at 3 , 8 , but the parabola 3 is always above the line otherwise. y 13 12 11 10 (+) 0 0 2 3 9 (+) 8 7 6 1 5 4 3 2 1 −1 1 x 4. To solve our last inequality, 2x...
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