Stitz-Zeager_College_Algebra_e-book

The second is that the function is broken up into two

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Unformatted text preview: ‘large’ denominator 1− 4x x−3 =0 1= (1)(x − 3) = 4x x−3 4x $ (x$$ clear denominators − 3) $3 $ x −$ $ x − 3 = 4x −3 = 3x −1 = x So we get two real numbers which make denominators 0, namely x = −1 and x = 3. Our domain is all real numbers except −1 and 3: (−∞, −1) ∪ (−1, 3) ∪ (3, ∞). 1 2 or, ‘implicit domain’ The word ‘implied’ is, well, implied. 48 Relations and Functions 2. The potential disaster for g is if the radicand3 is negative. To avoid this, we set 4 − 3x ≥ 0 4 − 3x ≥ 0 4 ≥ 3x 4 ≥x 3 Hence, as long as x ≤ 4 , the expression 4 − 3x ≥ 0, and the formula g (x) returns a real 3 4 number. Our domain is −∞, 3 . 3. The formula for h(x) is hauntingly close to that of g (x) with one key difference − whereas the expression for g (x) includes an even indexed root (namely a square root), the formula for h(x) involves an odd indexed root (the fifth root.) Since odd roots of real numbers (even negative real numbers) are real numbers, there is no restriction on the inputs to...
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