Pre-Calc Homework Solutions 172

1 l01 11 b at the given values dy 215 sec 152 1

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Unformatted text preview: .21 1x x 1 0.15 sec 1.25 tan 1.25 27. (a) f f (0.1) f (0) 0.21 (b) Since f (x) 2x 2, f (0) 2. Therefore, df 2 dx 2(0.1) 0.2. (c) 28. (a) f f df f (1.1) 0.21 f (1) 2 (c) The actual value is less than 1.1. This is because the derivative is decreasing over the interval [0, 0.1], which means that the graph of f (x) is concave down and lies below its linearization in this interval. 0.2 0.01 0 0.231 2. 0.2. 0.231 1, f (1) (b) Since f (x) 3x Therefore, df 2dx 2(0.1) (c) f df 0.231 0.2 0.031...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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