MATH 151 Midterm: MATH 151 TAMU Y2008 2008c Exam 2a Solutions
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Solutions-form a: c: 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) 3 (cid:19)(cid:18) sin(3 : d: first we will nd the tangent vector at t = 0 and then make it a unit vector by dividing by the magnitude: r r(t) = (cid:10)e2t, t cos t(cid:11) thus. (t) = (cid:10)2e2t, cos t t sin t(cid:11). (0) = (cid:10)2e0, cos(0) (0) sin(0)(cid:11) = h2, 1i. 5(cid:29): c: to solve ln(x + e) + ln(x e) = 2 + ln(3), we will. 3 (cid:19) = 2 (x + e)(x e) use logarithm properities: ln(x + e) + ln(x e) ln(3) = 2. (x + e)(x e) = 3e2. X2 e2 = 3e2, yielding x = 2e.
