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 ( 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 ( t ) = 2 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 ( z ) = 2 az + b = 2 z +25. C03S01.004: If f ( x ) = 49 x + 16, then a = 0, b = 49, and c = 16, so f ( x ) = 49. C03S01.005: If y = 2 x 2 + 3 x 17, then a = 2, b = 3, and c = 17, so dy dx = 2 ax + b = 4 x + 3. C03S01.006: If x = 100 t 2 + 16 t , then a = 100, b = 16, and c = 0, so dx dt = 2 at + b = 200 t + 16. C03S01.007: If z = 5 u 2 3 u , then a = 5, b = 3, and c = 0, so dz du = 2 au + b = 10 u 3. C03S01.008: If v = 5 y 2 + 500 y , then a = 5, b = 500, and c = 0, so dv dy = 2 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 = 2 ay + b = 10 y + 17. C03S01.010: If u = 7 t 2 + 13 t , then a = 7, b = 13, and c = 0, so du dt = 2 at + b = 14 t + 13. C03S01.011: f ( 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 ( 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 ( 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 + 2 xh + h 2 + 5 x 2 5 h = lim h 0 2 xh + h 2 h = lim h 0 (2 x + h ) = 2 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 ( x ) = lim h 0 f ( x + h ) f ( x ) h = lim h 0 1 2( x + h ) + 1 1 2 x + 1 h = lim h 0 2 x + 1 (2 x + 2 h + 1) h (2 x + 2 h + 1)(2 x + 1) = lim h 0 2 x + 1 2 x 2 h 1 h (2 x + 2 h + 1)(2 x + 1) = lim h 0 2 h h (2 x + 2 h + 1)(2 x + 1) 1
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= lim h 0 2 (2 x + 2 h + 1)(2 x + 1) = 2 (2 x + 1) 2 . C03S01.016: f ( 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 ( 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 )( 2 x + 2 h + 1 + 2 x + 1 ) h ( 2 x + 2 h + 1 + 2 x + 1 ) = lim h 0 (2 h + 2 h + 1) (2 x + 1) h ( 2 x + 2 h + 1 + 2 x + 1 ) = lim h 0 2 h h ( 2 x + 2 h + 1 + 2 x + 1 ) = lim h 0 2 2 x + 2 h + 1 + 2 x + 1 = 2 2 2 x + 1 = 1 2 x + 1 . C03S01.018: f ( x ) = lim h 0 1 h ( f ( x + h ) f ( x )) = lim h 0 1 h 1 x + h + 1 1 x + 1 = lim h 0 x + 1 x + h + 1 h x + h + 1 x + 1 = lim h 0 ( x + 1 x + h + 1 )( x + 1 + x + h + 1 ) h ( x + h + 1 x + 1 )( x + 1 + x + h + 1 ) = lim h 0 ( x + 1) ( x + h + 1) h ( x + h + 1 x + 1 )( x + 1 + x + h + 1 ) = lim h 0 h h ( x + h + 1 x + 1 )( x + 1 + x + h + 1 ) = lim h 0 1 ( x + h + 1 x + 1 )( x + 1 + x + h + 1 ) = 1 ( x + 1 ) 2 ( 2 x + 1 ) = 1 2( x + 1) 3 / 2 . C03S01.019: f ( x ) = lim h 0 1 h ( f ( x + h ) f ( x )) = lim h 0 1 h 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 + 2 x 2 + 2 xh h (1 2 x 2 h )(1 2 x ) = lim h 0 h h (1 2 x
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