Stitz-Zeager_College_Algebra_e-book

We are putting it on 117 by denition we get 6 in

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Unformatted text preview: /3 , we have ourselves a ‘quadratic in disguise’ and we can rewrite x4/3 − x2/3 − 6 = 0 2 as x2/3 − x2/3 − 6 = 0. If we let u = x2/3 , then in terms of u, we get u2 − u − 6 = 0. Solving for u, we obtain u = −2 or u = 3. Replacing x2/3 back in for u, we get x2/3 = −2 or x2/3 = 3. To avoid the trouble we encountered in the discussion√ following Definition 5.5, √ 3 3 we √ convert back to radical notation. By interpreting x2/3 as x2 we have x2 = −2 now 3 or x2 = 3. Cubing both sides of these equations results in x2 = −8, which admits no √ real solution, or x2 = 27, which √ gives x = ±3 3. We construct a sign diagram and find √ x4/3 − x2/3 − 6 > 0 on −∞, −3 3 ∪ 3 3, ∞ . To check our answer graphically, we set f (x) = x2/3 and g (x) = x4/3 − 6. The solution to x2/3 < x4/3 − 6 corresponds to the inequality f (x) < g (x), which means we are looking for the x values for which the graph of f is below the graph of g . Using the ‘Intersect’ command we...
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