MATH-M 211 Study Guide - Midterm Guide: The Roots, Pythagorean Theorem
Document Summary
Solutions to m211exam2 sp14 lim x 0 sin(2x) sin(8x) 4 sec x 1 x2 lim x 0. 1 x2 cos x 1 (cid:0) x cos x (sin x. For x = 0 and y = 1, (2) gives y(cid:48) = 1 whence y(cid:48)(cid:48) = 1. )y(cid:48)(cid:48) = 0: we have ln f (x) = x2 ln(cos(x)), so we di erentiate and obtain and therefore. = 2x ln(cos(x)) + x2 1 cos x ( sin x), 2x ln(cos(x)) x2 sin x cos x: a true, b false, c false, d true, f (x) = cos(e5x4 sin(e5x4 ) F (cid:48)(x) : j(cid:48)(x) is positive for x smaller than 1 or larger than 3 (approximately), and negative in between. So the slope of the tangent to the graph of j is negative for x approximately between 1 and 3 and negative otherwise. The only graph that has these properties is c. 7: the corresponding formula for the amount after t years is.
