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Unformatted text preview: Ma 115 Homework Solutions for Week 1 1.1 #23 f ( x ) = 4 + 3 x x 2 f (3 + h ) = 4 + 3(3 + h ) (3 + h ) 2 = 4 + 9 + 3 h (9 + 6 h + h 2 ) = 4 3 h h 2 f (3+ h ) f (3) h = ( 4 3 h h 2 4 ) h = h ( 3 h ) h = 3 h 1.1 #28 f ( x ) = 5 x +4 x 2 +3 x +2 is defined for all x except when x 2 + 3 x + 2 = 0 ( x + 1)( x + 2) = 0 x = 2 , 1 domain = { x R  x negationslash = 2 , 1 } = ( , 2) ( 2 , 1) ( 1 , ) 1.1 #32 h ( x ) = 4 x 2 . Now y = 4 x 2 y 2 = 4 x 2 x 2 + y 2 = 4, so the graph is the top half of a circle of radius 2 with center at the origin. The domain is { x  4 x 2 } = { x  4 x 2 } = { x  2  x } = [ 2 , 2]. The range is 0 y 2, or [0 , 2]. x y 2 2 2 1.1 #44 f ( x ) = 1 if x 1 3 x + 2 if 1 &lt; x &lt; 1 7 2 x if x 1 The domain is R ( 1 , 1) (1 , 5) y x 1 1.1 #54 Let the volume of the cube be V and the length of an edge be L . The V = L 3 so L = 3 V , and the surface area is S ( V ) = 6( 3 V ) 2 = 6 V 2 3 , with domain V &gt; 0....
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.
 Fall '08
 Mahalanobis

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