MATH 151 Midterm: MATH 151 TAMU Y2008 2008c Exam 2b Solutions
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Solutions-form b: c: let y = x = Solve for y: x(4y + 3) = 1 y. 4xy + y = 1 3x, hence y = e1/x = 0 since lim x 0 lim x 0 : e: e1/x = e(cid:18) lim x 0 and lim x 0 . 1/x(cid:19) = lim y ey = 0: b: the quadratic approximation for f (x) at x = 1 f (1) (x 1)2. 2 is q(x) = f (1) + f (1)(x 1) + f (1) = 2, f (1) = 2 and f (1) = 4. Q(x) = 2 2(x 1) + 2(x 1)2, and. 2 a(cid:18) : e: let f (x) = x5 2x + 5. Newton"s method says if x1 is the rst guess to f (x) = 0, then the second approximation is given by x2 = x1 if x1 = 1, x2 = 1 . = 1 4/3 = f (x1) f (x1) f (1) f (1)
