Pre-Calc Homework Solutions 170

Pre-Calc Homework Solutions 170 - 170 Section 4.5 1 2 x 2...

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Unformatted text preview: 170 Section 4.5 1 2 x 2 2. (a) f (x) (2x) 9 x x 2 8. (a) f (x) 9 4 . 5 (1 2 1 x x)6 [1 2[1 ( x)]6 ( x)] 1 1 2[1 6( x) 1 6x We have f ( 4) L(x) f ( 4) 5 4 x 5 4 (x 5 9 5 5 and f ( 4) f ( 4)(x 4) ( 4)) (b) f (x) 2 (c) f (x) (d) f (x) ( 1)( x)] 2x (1 2 2 1 x) 1/2 1 21 21 41/3 1 41/3 1 2/3 1 x 2 x 2 1/2 2 x2 4 3x 1/3 4 x 4 1 x 2 (b) Since f ( 3.9) 4.9204 and L( 3.9) 4.92, the approximation differs from the true value by less than 10 3. 3. (a) f (x) 1 x We have f (1) 2 and f (1) L(x) f (1) f (1)(x 1) 2 0(x 1) 2 2 x2 1 x2 2 2 0. (e) f (x) (4 3x)1/3 1 3x 3 4 1 2 1 2 x x 41/3 1 (f) f (x) 1 9. f (x) 1 2 3 1 2 x (b) Since f (1.1) 2.009 and L(1.1) 2, the approximation differs from the true value by less than 10 2. 4. (a) f (x) 1 x 1 1 1 2 6 3x 1 2 x 2/3 We have f (0) 0 and f (0) L(x) f (0) f (0)(x 0) 0 1x x 1. cos x 1 We have f (0) L(x) f (0) 1 3 x 2 1 and f (0) f (0)(x 0) 3 2 (b) Since f (0.1) 0.0953 and L(0.1) 0.1 the approximation differs from the true value by less than 10 2. 5. (a) f (x) sec2 x We have f ( ) 0 and f ( ) L(x) f ( ) f ( )(x ) 0 1(x ) x 1. The linearization is the sum of the two individual linearizations, which are x for sin x and 1 for x 1. 0.002)100 0.021 0.009)1/3 9 10 10 1 6 1 x 2 (b) Since f ( 0.1) 0.10033 and L( 0.1) 0.1, the approximation differs from the true value in absolute value by less than 10 3. 6. (a) f (x) 1 1 x2 2 10. (a) (1.002)100 (1 1.2; 1.002100 1.2 (b) 3 3 1 1 (100)(0.002) 1.009 1.009 (1 1 (0.009) 3 1.003; 1.003 10 5 We have f (0) L(x) f (0) 2 and f (0) 0) 0) 1. f (0)(x ( 1)(x 11. Center 1 f (x) 4x 4 We have f ( 1) 5 and f ( 1) 0 L(x) f ( 1) f ( 1)(x ( 1)) 5 12. Center f (x) 8 1 2/3 x 3 0(x 1) 5 x 2 (b) Since f (0.1) 1.47063 and L(0.1) 1.47080, the approximation differs from the true value in absolute value by less than 10 3. 7. f (x) k(1 x) We have f (0) 1 and f (0) L(x) f (0) f (0)(x 0) 1 k(x 0) 1 kx k 1 We have f (8) L(x) f (8) 2 and f (8) f (8)(x 8) 1 . 12 k. 2 1 (x 12 8) x 12 4 3 ...
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