Pre-Calc Homework Solutions 170

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

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2. (a) f 9 ( x ) 5 } 2 ˇ x 1 2 w 1 w 9 w } (2 x ) 5 } ˇ x 2 w x 1 w 9 w } We have f ( 2 4) 5 5 and f 9 ( 2 4) 5 2 } 4 5 } . L ( x ) 5 f ( 2 4) 1 f 9 ( 2 4)( x 2 ( 2 4)) 5 5 2 } 4 5 } ( x 1 4) 5 2 } 4 5 } x 1 } 9 5 } (b) Since f ( 2 3.9) < 4.9204 and L ( 2 3.9) 5 4.92, the approximation differs from the true value by less than 10 2 3 . 3. (a) f 9 ( x ) 5 1 2 x 2 2 We have f (1) 5 2 and f 9 (1) 5 0. L ( x ) 5 f (1) 1 f 9 (1)( x 2 1) 5 2 1 0( x 2 1) 5 2 (b) Since f (1.1) < 2.009 and L (1.1) 5 2, the approximation differs from the true value by less than 10 2 2 . 4. (a) f 9 ( x ) 5 } x 1 1 1 } We have f (0) 5 0 and f 9 (0) 5 1. L ( x ) 5 f (0) 1 f 9 (0)( x 2 0) 5 0 1 1 x 5 x (b) Since f (0.1) < 0.0953 and L (0.1) 5 0.1 the approximation differs from the true value by less than 10 2 2 . 5. (a) f 9 ( x ) 5 sec 2 x We have f ( p ) 5 0 and f 9 ( p ) 5 1. L ( x ) 5 f ( p ) 1 f 9 ( p )( x 2 p ) 5 0 1 1( x 2 p ) 5 x 2 p (b) Since f ( p 1 0.1) < 0.10033 and L ( p 1 0.1) 5 0.1, the approximation differs from the true value in absolute value by less than 10 2 3 . 6. (a) f 9 ( x ) 5 2 } ˇ 1 w 1 2 w x w 2 w } We have f (0) 5 } p 2 } and f 9 (0) 5 2 1. L ( x ) 5 f (0) 1 f 9 (0)( x 2 0) 5 } p 2 } 1 ( 2 1)( x 2 0) 5 2 x 1 } p 2 } (b) Since f (0.1) < 1.47063 and L (0.1) < 1.47080, the
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