Pre-Calc Homework Solutions 175

Pre-Calc Homework Solutions 175 - Section 4.6 (e) h(x) h...

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(e) h ( x ) 5 (1 1 x ) 1/2 h 9 ( x ) 5 } 1 2 } (1 1 x ) 2 1/2 h 0 ( x ) 52} 1 4 } (1 1 x ) 2 3/2 Since h (0) 5 1, h 9 (1) 5 } 1 2 } , and h 0 (1) 1 4 } , the coefficients are b 0 5 1, b 1 5 } 1 2 } , and b 2 55 2 } 1 8 } . The quadratic approximation is Q ( x ) 5 1 1 } 2 x } 2 } x 8 2 } . [ 2 1.35, 3.35] by [ 2 1.25, 3.25] As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical. (f) The linearization of any differentiable function u ( x ) at x 5 a is L ( x ) 5 u ( a ) 1 u 9 ( a )( x 2 a ) 5 b 0 1 b 1 ( x 2 a ), where b 0 and b 1 are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for f ( x ) at x 5 0 is 1 1 x ; the linearization for g ( x ) at x 5 1 is 1 2 ( x 2 1) or 2 2 x ; and the linearization for h ( x ) at x 5 0 is 1 1 } 2 x } . 53. Just multiply the corresponding derivative formulas by dx . (a) Since } d d x } ( c ) 5 0, d ( c ) 5 0. (b) Since } d d x } ( cu ) 5 c } d d u x } , d ( cu ) 5 c du . (c) Since } d d x } ( u + v ) 5 } d d u x } 1 } d d v x } , d ( u 1 v ) 5 du 1 dv (d) Since } d d x } ( u ? v ) 5 u } d d v x } 1 v } d d u x } , d ( u ?
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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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