Chapter 1 Answers Homework EVEN #

# Chapter 1 Answers Homework EVEN # - Chapter 1 Functions,...

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1 Chapter 1 Functions, Graphs, and Limits 1.1 Functions 2. f ( x ) = 7 x + 1 f (0) = 7(0) + 1 = 1 f (1) = 7(1) + 1 = 6 f ( 2) = 7( 2) + 1 = 15 4. 33 () (2 1 ) ( 1 ) ( 2 1 ) 1 ( 0 )( 01 ) 1 ( 1 21 ) 2 7 ht t h hh =+ = += =+ = 6. 22 2 () ( 2 ) 5 12 1 1 (0) 0 ( 1) 2 (1 ) 1 x fx f x ff == = ++ = = + −+ 8. 32 3 () ( 1 ) (0) (0 1) 1 ) (1 1 ) 0 (8) (8 1) 9 27 gu u g g g = −=−+ = = = 10. 1 1 (1) 1 ( 1 ) 11 (3 ) 3 ) (0) ( 0 ) 3 = −= = −− ft t f f f 12. () 4 (2 ) 4 2 6 (0) 4 0 4 (2) 4 2 6 gx x g g g −=+ 14. 3i f 5 1 if 5 5 if 5 <− ≤≤ > t t t tt f ( 6) = 3 f ( 5) = 5 + 1 = 4 (16) 16 4 f 16. Since 2 10 x for 1 x , f ( x ) is defined only for 1 x ≠± and the domain does not consist of the real numbers. 18. The square root function only makes sense for non-negative numbers. Since 2 t +≥ for all real numbers t the domain of 2 1 t consists of all real numbers. 20. 3 2 5 fx x x x =− The domain consists of all reals. 22. 2 2 1 2 2( 2 ) )0 t t t + = −− = − + ≠ if 1 and 2. ≠− 24. 2 4 hs s is defined only if 2 40 s −≥ or equivalently ) ) 0 ss . This occurs when the factors ) s and ) s + are zero or have the same sign. This happens when 2 s or 2 s ≤− and these values of s form the domain of h . 26. 2 4 ) )4 25 fu u x x x −= − += − +

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2 Chapter 1. Functions, Graphs, and Limits 28. 2 ( ) (2 10) fu u =+ [] 2 22 (5 )2 ) 1 0 (2 10 10) 4 fx x xx −= −+ =− + = 30. 2 2 1 () 1 (2 ) 2 u fx x = +− = +− 32. 2 2 11 , 1 (1 ) fu u f x x ⎛⎞ == ⎜⎟ ⎝⎠ 34. For () 2 3 x , ( ) ( 2 3 ) ( 2 3 ) 223 23 2 2 fx h fx x h x hh xh x h h h ++− + = ++ −− = = = 36. 2 = 2 ( ) 2 ) 2 + = = + = x h h x h hxh h 38. 1 = x ( ) 1 + + =⋅ + = + = + = + = + xh x xx h x x h xxh hx x h hx x h h hx x h 40. 2 1 , () 1 gx x = 2 2 2 2 (() ) ( 1 ) 1 a n d ) 1( 1 ) fgx x gfx x x + =−+ + So ( ( )) ( ( )) = means 2 2 2 0 10 x = = −+= but, by the quadratic formula, this last equation has no solutions. 42. 4 2 12 ) 4 x x x x + + , 1 1 4 41 ) 21 2 x x x x + + so ( ( )) ( ( )) = means 24 1 42 1 = . Clearing denominators, multiplying, and collecting terms gives 2 (2 )(2 1) (4 1)(4 ) 25 2 7 4 61 2 6 6 1 ) 6( 0 x x x x x x x = ++= − + − = + = The last equation has solution 1 x = which is in the domains of (() ) and ) . Thus ( ( )) ( ( )) = only for 1 x = . 44. 2 5 ) )5 26 x x x += + += + + 46. 2 () ( 2 6 ) x 2 2 2 (3 ) 6 6 6) 4 x x x += =
Chapter 1. Functions, Graphs, and Limits 3 48. 1 2 () 3 11 2 3 (3) 2 x fx x x xxx =+ ⎛⎞ = + ⎜⎟ ⎝⎠ 50. () 2 2 0 x =− 22 2 2 (2 9 ) 2 9 ) 2 0 2 4 18 20 24 2 x x x xx −+= −+ + 52. 52 3 3 () ( 3 1 2 ) () 3 1 2 x x gu u hx x x =− + = + 54. [] ( ) 3 5 can be written as ( ) with ( ) and ( ) 3 5. x ghx u x == 56. 33 4 . () (4 ) 4 x u xu x = − + 58. (a) 2 (0 . 3 7 4 7 ) 0 . 3 7 4 7 + = − + R D x x x 2 ( ) ( ) ( ) ( 0.37 47 ) (1.38 15.15 115.5) 1.75 31.85 115.5 =−= +− + + = + Px Rx Cx x x x x x x (b) Since 2 ( ) 1.75 31.85 115.5 x x + , the quadratic formula tells us () 0 = when 5 x = and 13.2 x = . By evaluating () at a number of x values, or by graphing the function, it is easy to see that () 0 > , that is the commodity is profitable, for 5 13.2 x << . 60. (a) 2 ( ) ( ) ( 0.09 51) 0.09 51 R D x x x + = 2 ( ) ( ) ( ) 0.09 51 (1.32 11.7 101.4) 1.41 39.3 101.4 x x x x x x + + + = + (b) Since 2 ( ) 1.41 39.3 101.4, x x + the quadratic formula tells us P ( x ) = 0 when x 2.8 and x 25.0. By evaluating P ( x ) at a number of x values, it is easy to see that P ( x ) > 0, that is, the commodity is profitable for 2.8 < x < 25.0.

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## This note was uploaded on 05/20/2011 for the course MAC 2233 taught by Professor Royer during the Spring '08 term at FIU.

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Chapter 1 Answers Homework EVEN # - Chapter 1 Functions,...

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