Stitz-Zeager_College_Algebra_e-book

1 338 exponential and logarithmic functions x we nd

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Unformatted text preview: no domain concerns there. However, the negative exponent on the second term indicates a denominator. x Rewriting r(x) with positive exponents, we obtain r(x) = 3(2 − x)1/3 − (2−x)2/3 . Setting the denominator equal to zero we get (2 − x)2/3 = 0, or 3 (2 − x)2 = 0. After cubing both sides, and subsequently taking square roots, we get 2 − x = 0, or x = 2. Hence, the domain of r is (−∞, 2) ∪ (2, ∞). To ﬁnd the zeros of r, we set r(x) = 0. There are two school of thought on how to proceed and we demonstrate both. • Factoring Approach. From r(x) = 3(2 − x)1/3 − x(2 − x)−2/3 , we note that the quantity (2 − x) is common to both terms. When we factor out common factors, we factor out the 2 1 quantity with the smaller exponent. In this case, since − 3 < 3 , we factor (2 − x)−2/3 from both quantities. While it may seem odd to do so, we need to factor (2 − x)−2/3 2 from (2 − x)1/3 , which results in subtracting the exponent − 3 from 1 . We proceed using...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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