Stitz-Zeager_College_Algebra_e-book

To check this answer analytically we rst check that f

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Unformatted text preview: ))(x) = 2−√x2 −4x+3 +1 . For the square root, we need ( ) x2 − 4x + 3 ≥ 0, which we determined in number 1 to be (−∞, 1] ∪ [3, ∞). Next, we set √ √ the denominator to zero and solve: 2 − x2 − 4x + 3 + 1 = 0. We get x2 − 4x + 3 = 3, So we get (h ◦ (g ◦ f ))(x) = and, after squaring both sides, we have √ − 4x + 3 = 9. To solve x2 − 4x − 6 = 0, we use x2 the quadratic formula and get x = 2 ± 10. The reader is encouraged to check that both √ of these numbers satisfy the original equation, 2 − x2 − 4x + 3 + 1 = 0. Hence we must exclude these numbers from the domain of h ◦ (g ◦ f ). Our final domain for h ◦ (f ◦ g ) is √ √ √ √ (−∞, 2 − 10) ∪ (2 − 10, 1] ∪ 3, 2 + 10 ∪ 2 + 10, ∞ . 7. The expression ((h ◦ g ) ◦ f )(x) indicates that we first find the composite h ◦ g and then compose √ x+3 that with f . From number 4, we gave (h ◦ g )(x) = 43−2√x+3 . We now proceed as before. − • inside out : We insert the expression f (x) into h ◦ g first to get ((h ◦ g ) ◦ f )(x) = = = = (h ◦ g )(f (x)) (h ◦ g ) x2 − 4x 4 − 2 (...
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