Calculus with Analytic Geometry by edwards & Penney soln ch3

# Calculus with Analytic Geometry

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Section 3.1 C03S01.001: Given f ( x )=4 x 5, we have a =0, b = 4, and c = 5, so f 0 ( x )=2 ax + b =4. C03S01.002: Given g ( t )= 16 t 2 + 100, we have a = 16, b = 0, and c = 100, so g 0 ( t at + b = 32 t . C03S01.003: If h ( z z (25 z z 2 +25 z , then a = 1, b = 25, and c =0,so h 0 ( z az + b = 2 z +25. C03S01.004: If f ( x 49 x + 16, then a b = 49, and c = 16, so f 0 ( x 49. C03S01.005: If y =2 x 2 +3 x 17, then a =2, b = 3, and c = 17, so dy dx ax + b =4 x +3. C03S01.006: If x = 100 t 2 +16 t , then a = 100, b = 16, and c dx dt at + b = 200 t + 16. C03S01.007: If z =5 u 2 3 u , then a =5, b = 3, and c dz du au + b =10 u 3. C03S01.008: If v = 5 y 2 + 500 y , then a = 5, b = 500, and c dv dy ay + b = 10 y + 500. C03S01.009: If x = 5 y 2 +17 y + 300, then a = 5, b = 17, and c = 300, so dx dy ay + b = 10 y + 17. C03S01.010: If u =7 t 2 +13 t , then a =7, b = 13, and c du dt at + b =14 t +13. C03S01.011: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 2( x + h ) 1 (2 x 1) h = lim h 0 2 x +2 h 1 2 x +1 h = lim h 0 2 h h = lim h 0 2=2. C03S01.012: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 2 3( x + h ) (2 3 x ) h = lim h 0 2 3 x 3 h 2+3 x h = lim h 0 3 h h = lim h 0 ( 3) = 3. C03S01.013: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 ( x + h ) 2 +5 ( x 2 +5) h = lim h 0 x 2 xh + h 2 x 2 5 h = lim h 0 2 xh + h 2 h = lim h 0 (2 x + h x . C03S01.014: f ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 3 2( x + h ) 2 (3 2 x 2 ) h = lim h 0 3 2 x 2 4 xh 2 h 2 3+2 x 2 h = lim h 0 4 xh 2 h 2 h = lim h 0 ( 4 x 2 h 4 x . C03S01.015: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 2( x + h )+1 1 2 x h = lim h 0 2 x (2 x h +1) h (2 x h + 1)(2 x = lim h 0 2 x 2 x 2 h 1 h (2 x h + 1)(2 x = lim h 0 2 h h (2 x h + 1)(2 x 1

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= lim h 0 2 (2 x +2 h + 1)(2 x +1) = 2 (2 x 2 . C03S01.016: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 3 ( x + h ) 1 3 x h = lim h 0 (3 x ) (3 x h ) h (3 x h )(3 x ) = lim h 0 3 x 3+ x + h h (3 x h )(3 x ) = lim h 0 h h (3 x h )(3 x ) = lim h 0 1 (3 x h )(3 x ) = 1 (3 x ) 2 . C03S01.017: f 0 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 p 2( x + h )+1 2 x +1 h = lim h 0 ( 2 x h 2 x )( 2 x h +1+ 2 x ) h ( 2 x h 2 x ) = lim h 0 (2 h h (2 x h ( 2 x h 2 x ) = lim h 0 2 h h ( 2 x h 2 x ) = lim h 0 2 2 x h 2 x = 2 2 2 x = 1 2 x . C03S01.018: f 0 ( x ) = lim h 0 1 h ( f ( x + h ) f ( x )) = lim h 0 1 h µ 1 x + h 1 x = lim h 0 x x + h h x + h x = lim h 0 ( x x + h x x + h ) h ( x + h x x x + h ) = lim h 0 ( x ( x + h h ( x + h x x x + h ) = lim h 0 h h ( x + h x x x + h ) = lim h 0 1 ( x + h x x x + h ) = 1 ( x ) 2 ( 2 x ) = 1 2( x 3 / 2 . C03S01.019: f 0 ( x ) = lim h 0 1 h ( f ( x + h ) f ( x )) = lim h 0 1 h µ x + h 1 2( x + h ) x 1 2 x = lim h 0 1 h · ( x + h )(1 2 x ) (1 2 x 2 h )( x ) (1 2 x 2 h )(1 2 x ) = lim h 0 ( x 2 x 2 + h 2 xh ) ( x 2 x 2 2 xh ) h (1 2 x 2 h )(1 2 x ) = lim h 0 x 2 x 2 + h 2 xh x x 2 xh h (1 2 x 2 h )(1 2 x ) = lim h 0 h h (1 2 x 2 h )(1 2 x ) = lim h 0 1 (1 2 x 2 h )(1 2 x ) = 1 (1 2 x ) 2 .
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## This document was uploaded on 02/01/2008.

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Calculus with Analytic Geometry by edwards &amp; Penney...

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