MATH118FS06S - Math 118 Final Exam Answeres Spring 2006 Page 1 of 2 1 a n=0 dy dx xn R=1 b = dy dt dx dt c f(x = n=0 f(n(a(x n a)n d p > 1 2 a ex C b ln

MATH118FS06S - Math 118 Final Exam Answeres Spring 2006...

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Math 118 - Final Exam Answeres Spring 2006 Page 1 of 2 1. a) n =0 x n , R = 1 b) dy dx = dy dt dx dt c) f ( x ) = n =0 f ( n ) ( a ) n ! ( x - a ) n d) p > 1 2. a) - e - x 2 + C b) ln x - 1 x - 2 + C c) 8 sin - 1 ( x 4 ) + x 2 16 - x 2 + C d) The improper integral diverges. 3. a) b) The slope of the tangent line is -1. 4. a) 1 2 b) Prove by induction the sequence is increasing. Prove that 1 is a lower bound and 3 is an upper bound. The limit is 3. 5. a) This is a geometric series with r = 1 e < 1 hence it converges. b) By the alternating series test, the series converges. c) The sereies converges by the comparison test. d) The series diverges by the n -th term test. 6. The interval of convergence is [ - 3 , 1).
Math 118 - Final Exam Answeres Spring 2006 Page 2 of 2 7.The Taylor Polynomial that approximatesf(x) on [0,4] with error less than 0.1 is4en! 2 P4(x) =n=0n(x-2)n=e+e2(x-2) +e8(x-2)2+e48(x-2)3+e384(x-2)48.1/3cos(x)dx13-136=1136with error less than 0.001 since 72·27>1000.9.n=0n+1n!=e1+ 1·e1= 2e10.a)y=Cex2-1b)y=6x(x2+1)2c)2DDy-1 =±x+Fd)y(0) =e2e2+e2x211.204 grams of salt.

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