MATH 151 Midterm: MATH 151 TAMU Y2011 2011a Exam 3a Solutions

• School: Texas A&M University
• Department: Mathematics
• Course: MATH 151
• Professor: All

Document Summary

Exam iii version a solutions: d since the numerator and denomina- tor both approach 0, use l"hospital"s rule: lim x 0 ex cos x 2x x2 2x. 2: e use properties of logarithms: ln(x2 + x) = ln(x + 4), so x2 + x = x + 4, x2 4 = 0, which yields x = 2 or x = 2. Therefore, f has a local minimum only when x = c: e let y = log4(cid:18) 1. 2 3, which means 2y = 3 and y = (alternatively, log4(cid:18) 1. 3 and cancel: a exponential growth means y = cekt, where y is the population after t minutes. When t = 0, y = 200, so y = 200ekt. When t = 30, y = 600, so 3 = ek(30), 30k = ln 3, or. When t = 45, y = 200eln 3/30 45 = k =