Stitz-Zeager_College_Algebra_e-book

Definition 18 suppose f is a function dened on an

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Unformatted text preview: (Why?) Graphically, The graph of h appears to be neither symmetric about the y -axis nor the origin. 4. i(x) = i(−x) = i(−x) = 5x 2x − x3 5(−x) 2(−x) − (−x)3 −5x −2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in fact, equal i(x). However, while this suggests i is even, it doesn’t prove it. (It does, however, 1.7 Graphs of Functions 69 prove i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x) and i(−x) and show they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3 ) −5x −2x + x3 Hence, i(x) = i...
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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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