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Pre-Calc Homework Solutions 11

Pre-Calc Homework Solutions 11 - Section 1.2 55 58 11 5 5...

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55. [ 2 5, 5] by [ 2 2, 5] We require x 2 2 4 $ 0 (so that the square root is defined) and x 2 2 4 0 (to avoid division by zero), so the domain is ( 2‘ , 2 2) < (2, ). For values of x in the domain, x 2 2 4 1 and hence ˇ x 2 w 2 w 4 w and } ˇ x 2 w 1 2 w 4 w } 2 can attain any positive value, so the range is (0, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher. 56. [ 2 5, 5] by [ 2 2, 5] We require 9 2 x 2 $ 0 (so that the fourth root is defined) and 9 2 x 2 0 (to avoid division by zero), so the domain is ( 2 3, 3). For values of x in the domain, 9 2 x 2 can attain any value in (0, 9]. Therefore, ˇ 4 9 w 2 w x w 2 w can attain any value in (0, ˇ 3 w ], and } ˇ 4 9 w 2 2 w x w 2 w } can attain any value in 3 } ˇ 2 3 w } , 2 . The range is 3 } ˇ 2 3 w } , 2 or approximately [1.15, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher.) 57. [ 2 4.7, 4.7] by [ 2 3.1, 3.1] We require 9 2 x 2 0, so the domain is ( 2‘ , 2 3) < ( 2 3, 3) < (3, ). For values of x in the domain, 9 2 x 2 can attain any value in ( 2‘ , 0) < (0, 9], so ˇ 3 9 w 2 w x w 2 w can attain any value in ( 2‘ , 0) < (0, ˇ 3 9 w ]. Therefore, } ˇ 3 9 w 2 2 w x w 2 w } can attain any
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