{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

55. [ 2 5, 5] by [ 2 2, 5] We require x 2 2 4 \$ 0 (so that the square root is defined) and x 2 2 4 0 (to avoid division by zero), so the domain is ( 2‘ , 2 2) < (2, ). For values of x in the domain, x 2 2 4 1 and hence ˇ x 2 w 2 w 4 w and } ˇ x 2 w 1 2 w 4 w } 2 can attain any positive value, so the range is (0, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher. 56. [ 2 5, 5] by [ 2 2, 5] We require 9 2 x 2 \$ 0 (so that the fourth root is defined) and 9 2 x 2 0 (to avoid division by zero), so the domain is ( 2 3, 3). For values of x in the domain, 9 2 x 2 can attain any value in (0, 9]. Therefore, ˇ 4 9 w 2 w x w 2 w can attain any value in (0, ˇ 3 w ], and } ˇ 4 9 w 2 2 w x w 2 w } can attain any value in 3 } ˇ 2 3 w } , 2 . The range is 3 } ˇ 2 3 w } , 2 or approximately [1.15, ). (Note that grapher failure may cause the range to appear as a finite interval on a grapher.) 57. [ 2 4.7, 4.7] by [ 2 3.1, 3.1] We require 9 2 x 2 0, so the domain is ( 2‘ , 2 3) < ( 2 3, 3) < (3, ). For values of x in the domain, 9 2 x 2 can attain any value in ( 2‘ , 0) < (0, 9], so ˇ 3 9 w 2 w x w 2 w can attain any value in ( 2‘ , 0) < (0, ˇ 3 9 w ]. Therefore, } ˇ 3 9 w 2 2 w x w 2 w } can attain any value in ( 2‘ , 0) < 3 } ˇ 3 2 9 w } , 2 . The range is ( 2‘ , 0) < 3 } ˇ 3 2 9 w } , 2 or approximately ( 2‘ , 0) < [0.96, ). (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.) 58. [ 2 2.35, 2.35] by [ 2 1.55, 1.55] We require x 2 2 1 0, so the domain is ( 2‘ , 2 1) < ( 2 1, 1) < (1, ). For values of x in the domain, x 2 2 1 can attain any value in [ 2 1, 0) < (0, ), so ˇ 3 x 2 w 2 w 1 w can also attain any value in [ 2 1, 0) < (0, ). Therefore, } ˇ 3 x 1 w 2 w 1 w } can attain any value in ( 2‘ , 2 1] < (0, ). The range is ( 2‘ , 2 1] < (0, ). (Note that grapher failure can cause the intervals in the range to appear as finite intervals on a grapher.) 59. (a) (b) 60. (a) y x 2 2 1 –1 –1 –2 –2 y x 1.5 2 0 –2 –1.5 y x 1.5 2 0 –2 –1.5 Section 1.2 11

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

View Full Document
60. continued (b) 61. (a) (b) 62. (a) (b) 63. (a) Since ( f + g )( x ) 5 ˇ g w ( x w ) w 2 w 5 w 5 ˇ x 2 w 2 w 5 w , g ( x ) 5 x 2 . (b) Since ( f + g )( x ) 5 1 1 } g ( 1 x ) } 5 x , we know that } g ( 1 x ) } 5 x 2 1, so g ( x ) 5 } x 2 1 1 } . (c) Since ( f + g )( x ) 5 f 1 } 1 x } 2 5 x , f ( x ) 5 } 1 x } . (d) Since ( f + g )( x ) 5 f ( ˇ x w ) 5 ) x ) , f ( x ) 5 x 2 . The completed table is shown. Note that the absolute value sign in part (d) is optional. 64. (a) Note that the data in the table begins at x 5 20. (We do not include the initial investment in the data.) The power regression equation is y 5 27.1094 x 2.651044 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern