315_James Stewart Calculus 5 Edition Answers

315_James Stewart - Stewart Calculus ET 5e 0534393217;3 Differentiation Rules 3.11 Linear Approximations and Differentials 21(t 2030 130 21 P(2040

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Unformatted text preview: Stewart Calculus ET 5e 0534393217;3. Differentiation Rules; 3.11 Linear Approximations and Differentials 21 (t 2030) 130 21 P(2040) =21+ (2040 2030) 22.6% 130 21 P(2050) =21+ (2050 2030) 24.2% 130 21+ These predictions are probably too high since the tangent line lies above the graph at t =2030. N (1980) N (1985) 15.0 17.0 N (1990) N (1985) 19.3 17.0 = =0.4 and B= = =0.46 . Then 1980 1985 5 1990 1985 5 N (t ) N (1985) A+ B / N (1985)= lim =0.43 million / year. So t 1985 2 t 1985 4. Let A= / N (1984) N (1985)+N (1985)(1984 1985) 17.0+0.43( 1)=16.57 million. N (1995) N (2000) 22.0 24.9 / N (2000) = =0.58 million / year. 1995 2000 5 / N (2006) N (2000)+N (2000)(2006 2000) 24.9+0.58(6)=28.38 million. 5. f (x)=x 3 / 2 / / f (x)=3x , so f (1)=1 and f (1)=3 . With a=1 , L(x)= f (a)+ f (a)(x a) becomes / L(x)= f (1)+ f (1)(x 1)=1+3(x 1)=3x 2 . / / 7. f (x)=cos x L(x)= f f (x)= sin x , so f +f 2 3 / / f (x)=1/x , so f (1)=0 and f (1)=1 . Thus, L(x)= f (1)+ f (1)(x 1)=0+1(x 1)=x 1. 6. f (x)=ln x 1/3 8. f (x)= x =x / / x 2 / f ( x )= 1 x 3 L(x)= f ( 8)+ f ( 8)(x+8)= 2+ 2 2/3 2 =0 1 =0 and f x 2 / = 1 . Thus, 2 = x+ / 2 , so f ( 8)= 2 and f ( 8)= . 1 . Thus, 12 1 1 4 . (x+8)= x 12 12 3 9. f (x)= 1 x 2 ...
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This note was uploaded on 08/02/2011 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at University of California, Berkeley.

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