MATH 151 Midterm: MATH 151 TAMU Y2010 2010c Exam 3a Solutions
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Solutions-form a f (x) changes concavity when f (x) goes from increasing to decreasing, that is x = b and x = e. t . 2(cid:17) = 0 and lim since cos(cid:16) : d. if it is given that f (x) = 4 x 0+ cos t(cid:19) = , ln(x) = . x2 + 1. Take an antiderivative of f (x) to obtain f (x): f (x) = 4x 5 arctan x + c. Since f (1) = 0, 4(1) 5 arctan(1) + c = 0. 4 5(cid:16) f (x) = 4x 5 arctan x + 4(cid:17) + c = 0, yielding c = Thus: d. find lim x 0 (1 4x)1/x. This is an indeterminate form of the type 1 . Then ln y = ln (1 4x)1/x ln(1 4x) 1 x x ln(1 4x) ln y = lim x 0 lim x 0 is indeterminate form, x.
