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Math 125 (CRN 10240,11371,15269) Fall 2013 Quiz 8 (9-11-13) Carter Name 1. Determine the points at which the function is continuous. Remember to exclude certain values of x if necessary. x tan (x) y= 2 x +1 Solution. Since 0 < x2 + 1 there is no chance that the denominator is 0. Meanwhile, the numerator is undefined for any odd multiple of π/2. So the function is continuous as long as x 6= (2n + 1)π/2 for n = 0, ±1, ±2, ±3, . . .. 2. Evaluate the function f (x) = x3 − 15x + 1 at the points x = −4, x = 0, x = 1, and x = 4. Give a clear, concise, explanation that the equation x3 − 15x + 1 = 0 has three solutions in the closed interval [−4, 4]. Solution. f (−4) = −64 − 15(−4) + 1 = −64 + 60 + 1 = −3. f (0) = 1. f (1) = 13 − 15 · 1 + 1. f (4) = 43 − 15 · 4 + 1 = 64 − 60 + 1 = 5. Since f (x) is continuous and the signs of its values alternate from left to right as −, +, −, + the Intermediate Value Theorem (IVT) give roots in each of the intervals [−4, 0], [0, 1], and [1, 4].