Stitz-Zeager_College_Algebra_e-book

The exponential model didnt t the data as well as the

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Unformatted text preview: nd the zeros of r, we set r(x) = ln(x)+1 = 0 so that e e ln(x) = 0, and we find x = e0 = 1. In order to determine test values for r without resorting to the calculator, we need to find numbers between 0, 1 , and 1 which have a base of e. Since e 1 e ≈ 2.718 > 1, 0 < e1 < 1 < √e < 1 < e. To determine the sign of r e1 , we use the fact that 2 2 e ln e1 = ln e−2 = −2, and find r e1 = −−2 = 2, which is (+). The rest of the test values 2 2 2+1 are determined similarly. From our sign diagram, we find the solution to be 0, 1 ∪ [1, ∞). e 1 Graphing f (x) = ln(x)+1 and g (x) = 1, we see the the graph of f is below the graph of g on the solution intervals, and that the graphs intersect at x = 1. (+) 0 (−) 0 (+) 1 e 1 y = f (x) = 1 ln(x)+1 and y = g (x) = 1 2. Moving all of the nonzero terms of (log2 (x))2 < 2 log2 (x) + 3 to one side of the inequality, we have (log2 (x))2 − 2 log2 (x) − 3 < 0. Defining r(x) = (log2 (x))2 − 2 log2 (x) − 3, we get the domain of r is (0, ∞), due to the presence of the logarithm. To find the zeros of r, we set r(x) = (log2 (x))2 − 2 log2 (x) − 3 = 0 which r...
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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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