Pre-Calc Homework Solutions 59

Pre-Calc Homework Solutions 59 - Section 2.4 (b) 59 49. For...

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(b) [0, 24] by [0, 9] This is continuous for all values of x in the domain [0, 24] except for x 5 0, 1, 2, 3, 4, 5, 6. 45. (a) The function is defined when 1 1 } 1 x } . 0, that is, on ( 2‘ , 2 1) < (0, ). (It can be argued that the domain should also include certain values in the interval ( 2 1, 0), namely, those rational numbers that have odd denominators when expressed in lowest terms.) (b) [ 2 5, 5] by [ 2 3, 10] (c) If we attempt to evaluate f ( x ) at these values, we obtain f ( 2 1) 5 1 1 1 } 2 1 1 } 2 2 1 5 0 2 1 5 } 1 0 } (undefined) and f (0) 5 1 1 1 } 1 0 } 2 0 (undefined). Since f is undefined at these values due to division by zero, both values are points of discontinuity. (d) The discontinuity at x 5 0 is removable because the right-hand limit is 0. The discontinuity at x 52 1 is not removable because it is an infinite discontinuity. (e) [0, 20] by [0, 3] The limit is about 2.718, or e . 46.
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