Homework1

Homework1 - Ma 115 Homework Solutions for Week 1 1.1 #23 f...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 < x < 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 > 0....
View Full Document

This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.

Page1 / 6

Homework1 - Ma 115 Homework Solutions for Week 1 1.1 #23 f...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online