Pre-Calc Homework Solutions 455

# Pre-Calc Homework Solutions 455 - Appendix A3 x 4 x x2 4 1...

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Unformatted text preview: Appendix A3 x 4 x x2 4 1 ( 1)2 4 1 3 x x2 x 4 1 3 455 12. (a) lim 2 x 1 x 13. Step 1: For x 1 1, x 2 x2 1 1 x x 1 1 1 1 x 1 1 x2 1 (b) Step 1: 0.1 1 3 0.1 Step 2: x 52 for x near 1. 30 13 2 52 x x , 30 30 1 1 x 1. 1 1 1 , 1 1 1 1. 1}, , 1 near x 1. 0.1 13 7 30 30 x2 4 28 13 2 7 2 x x x 30 30 30 28 7 Then x 2 x 0 30 30 1 Then or which using the quadratic formula implies x x 15 13 15 7 15 7 901 421 421 0.7883 or 5.0740, and also 1.1551 15 13 Choose min {1 that is, the smaller of the two distances. 901 13 15 7 421 x 901 15 14. Step 1: . 1 x2 1 3 1 3 1 3 3 1 3 1 3 3 1 x2 1 x2 1 3 1 3 1 x2 1 3 3.4628. Thus Step 2: x Then or ( 1) 1 1 x 1 or Step 2: x x 15 13 x 1. 901 3 3 3 3 3 1 3 x2 3 1 3 x x x 1 1 3 3 , for x near 3 3. 901 28 0.1551, 13 15 421 1 7 22 421 0.2117. 7 3 3 3 3 x 3 3 1 3 3 3 . 1 Then or 3 3 3 3 1 3 , Choose Alternately, graph y1 y3 x x 1 3 0.155. x x2 4 1 3. , y2 1 3 0.1, and Choose min 15. (a) (5 I x5 0.1. The curve intersects the lines at 1 1 0.15513 and at 0.21167. We may choose (b) lim x 3 2 3 1 3 , 3 1 3 2 3 . 1.15513 0.78833 0.155. ) 5 (5, 5 5 ) 2 ) 0 , 4) 0 L, show that x c 2 16. (a) 4 I x4 (4 (4 4 x (b) lim [ 1.5, 0] by [ 0.15, 0.55] 17. If L, c, and k are real numbers and lim f (x) xc for any k f (x) 0, there is a k L . 0, let x x c 0 such that 0 Proof: For any is a k k . Since lim f (x) xc L, there 0 such that 0 . Therefore, 0 c f (x) k f (x) L k L . ...
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