Math141Sols1BF11

Math141Sols1BF11 - QUIZ 1B - SOLUTIONS (CHAPTER 1) MATH 141...

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QUIZ 1B - SOLUTIONS (CHAPTER 1) MATH 141 – FALL 2011 – KUNIYUKI 60 POINTS TOTAL No notes or books allowed. A scientific calculator is allowed. You may assume that two-dimensional graphs are in the usual Cartesian xy -plane. Give exact answers, unless you are told to approximate. SHORTER PROBLEMS (21 POINTS) 1) (6 points). Write the domain of f , where fr () = r ± 5 r ± 7 + r 3 , using interval form (the form using parentheses and/or brackets). The domain of 3 is ± , so it imposes no restrictions. is real ± r ² 5 ³ 0 and r ² 7 ´ 0 ± r ³ 5 and r ´ 7 Here is a graph of the domain: In interval form, we have: 5, 7 ± ² ) ³ 7, ´ 2) (3 points). The graph of y = 2 x 5 + 3 x 3 ± x is symmetric about the …. (Box in one:) x -axis y -axis origin (none of these) The function f , where fx = 2 x 5 + 3 x 3 ± x , is odd, because, for all x in the domain of f (which is ± ) , f ± x = 2 ± x 5 + 3 ± x 3 ±± x = ± 2 x 5 ± 3 x 3 + x = ± 2 x 5 + 3 x 3 ± x = ± In general, a polynomial in x where all exponents on x are odd (and that has no nonzero constant terms) represents an odd function of x .
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Here is the graph of f : 3) (3 points). Find functions g and f such that f ± g () x = 4 x + 5 3 . You may not use the identity function. Fill in the blanks: gx = 4 x + 5 fu = u 3 There are many other possibilities.
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This note was uploaded on 02/07/2012 for the course MATH 2203 taught by Professor Ellermeyer during the Fall '10 term at FIU.

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Math141Sols1BF11 - QUIZ 1B - SOLUTIONS (CHAPTER 1) MATH 141...

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