CK-12 PreCalculus Concepts 1

Transcription

1 Chapter Functions and Graphs Answer Ke. Functions Families CK- PreCalculus Concepts

2 Chapter Functions and Graphs Answer Ke CK- PreCalculus Concepts

3 Chapter Functions and Graphs Answer Ke CK- PreCalculus Concepts

4 Chapter Functions and Graphs Answer Ke =, because is undefined. 0. = e, =, =, =. One difference is = has a minimum value, while = doesn t.. The two graphs are reflections of one another across the line =. 5. = is not defined for all values of because the square root of an negative number is not a real number. CK- PreCalculus Concepts

5 Chapter Functions and Graphs Answer Ke. Graphical Transformations. Reflection across the ais and reflection across the ais.. Reflection across the ais and a horizontal shift left units.. Horizontal shift left unit and vertical shift down units.. Reflection across the ais and horizontal shift right units. 5. Reflection across the ais and horizontal compression b a factor of. 6. Vertical stretch b a factor of, horizontal stretch b a factor of, and horizontal shift left units. 7. A reflection across the ais, a horizontal shift right units, vertical shift down units, and a vertical stretch b a factor of. 8. Vertical stretch b a factor of 5 and a horizontal shift left unit. 9. h( ) + 0. f( + ). g( ). j(( )) +. k ( ( + )) +. h( ( )) 5. 5f() CK- PreCalculus Concepts 5

6 Chapter Functions and Graphs Answer Ke. Point Notation and Function Notation. Vertical reflection across the ais, vertical compression b a factor of, horizontal shift one unit left. (, ) (, ) Vertical stretch b a factor of, horizontal compression b a factor of, and vertical shift up units. (, ) (, + ) / / 6. Reflection across the ais, horizontal shift units to the right, vertical shift units down. (, ) ( +, ) Vertical stretch b a factor of, horizontal compression b a factor of, horizontal shift units to the right, and vertical shift up unit. (, ) ( +, + ) CK- PreCalculus Concepts 6

7 Chapter Functions and Graphs Answer Ke 5. Reflection across the ais, horizontal shift right units. (, ) ( +, ) f() f( 6) 7. f() f ( ) + 8. f() f ( ) 5 9. f() f ( ) + 0. f() f ( ) +. (, ) ( +, + ). (, ) ( +, + ). (, ) (, 5). (, ) ( +, 5 ) 5. (, ) ( +, ) CK- PreCalculus Concepts 7

8 Chapter Functions and Graphs Answer Ke. Domain and Range. (0, ). [,). [,) (,) (,5]. (5, ) 5. (, 0) (0, ) 6. (, 5] 7. (0, ) 8. (, ] (5, ) 9. Domain: (, ) Range: [,] 0. Domain: (, ) [, ) Range: [ 8, ). var. Domain should be [0, ) and range should be (,].. var. Domain should be [,) (, ) and range should be (, ).. Domain: {,, π,, } Range: {, π, 5, 7}. Domain: [, ) 5. Domain: (, 6) ( 6, ) 6. Domain: (, ) (, ) CK- PreCalculus Concepts 8

9 Chapter Functions and Graphs Answer Ke.5 Maimums and Minimums. There is a global minimum at (,0).. There are no local etrema.. Global minimum at (-π/, -) and global maimum at (π/, ).. Local maimum at (-π, 0) and global minimum at (π, 0). 5. There are no global etrema. 6. There are no local etrema. 7. There are no global etrema. 8. Local minimums: (0., -), (.5, -). Local maimums: (-.5, ), (, 0). Note: points are approimate. 9. There are no global etrema. 0. Local minimum: (, 0). Local maimum: (0.5, 9.5). Note: points are approimate.. A global maimum is the overall highest point on the graph, while the local maimum is the highest point within a certain neighborhood of the graph.. var. Graph should show a global minimum, a local maimum, no global maimum (there can be a local minimum).. var. Graph should have no global etrema, but both tpes of local etrema.. Local maimum: (-.6, 6.). Local minimum: (-, 0). No global maimum. Global minimum: (,6, -8.9). 5. Local maimum: (-,0). Local minimum: (0., -.). Global maimum: (.8, 9.9). No global minimum. CK- PreCalculus Concepts 9

10 Chapter Functions and Graphs Answer Ke.6 Smmetr. Even. Odd. Neither. Neither 5. Odd 6. Neither 7. Neither 8. Even 9. f( ) = h() + g() 0. f( ) = h() g(). f( ) = h()g(). Yes. If h() and g() are both even and f()=h()+g(), then: f( ) = h( ) + g( ) = h() + g() = f().. Yes. If h() and g() are both odd and f()=h()+g(), then: f( ) = h( ) + g( ) = h() g() = [h() + g()] = f().. There are some functions that do not have reflection smmetr across the ais or rotation smmetr about the origin. 5. If a function is even then it is smmetrical across the ais. If a function is odd then it has rotation smmetr about the origin. CK- PreCalculus Concepts 0

11 Chapter Functions and Graphs Answer Ke.7 Increasing and Decreasing. (, ). (, ). ( π, π ). ( π, π ) (π, π) 5. (, ) 6. None 7. (,.) (0., ) (.5, ) Note that points are approimate 8. (., 0.) (,.5) Note that points are approimate 9. (, 0.) (, ) 0. (0., ). var. Possible answer: A line with a positive slope.. var. Possible answer: A line with a negative slope.. Increasing: (, ) (, ). Decreasing: (, ).. Increasing: (, ). Decreasing: (, ). 5. Increasing: (5, ). Decreasing: (, 5) CK- PreCalculus Concepts

12 Chapter Functions and Graphs Answer Ke.8 Zeroes and Y-Intercepts of Functions. -intercept: (0, -); Zeroes: (-,0) and (0, -). -intercept: (0, -); Roots: (-,0), (,0) and (, 0). -intercept is approimatel (0, 5), -intercepts are (-,0) and (, 0). Both and intercepts are (0,0) 5. Both and intercepts are (0, 0) 6. No -intercept; -intercept is (, 0) 7. No or - intercepts 8. -intercept is (0, ); no -intercept 9. Both and -intercepts are (0,0) 0. Yes, because there are functions that are undefined when =0.. Yes, because there are functions with no real solutions when =0.. The -intercept of f() is called a zero because it is the solution to f()=0.. -intercept: (0, 0); -intercepts: (,0), (-,0), (5,0). -intercept: (0, -7); -intercepts: (-,0),(7,0) 5. -intercept: (0, 5); -intercepts: (5,0), (-/, 0), (,0) CK- PreCalculus Concepts

13 Chapter Functions and Graphs Answer Ke.9 Asmptotes and End Behavior. There are no asmptotes. As approaches positive infinit, approaches positive infinit. As approaches negative infinit, approaches negative infinit.. There are no asmptotes. As approaches both positive and negative infinit, approaches positive infinit.. There are no asmptotes. As approaches positive infinit, approaches positive infinit. As approaches negative infinit, approaches negative infinit.. There are no asmptotes. As approaches positive infinit, approaches positive infinit. 5. There is a horizontal asmptote at =0 and a vertical asmptote as =0. As approaches both positive and negative infinit, approaches As approaches negative infinit there is a horizontal asmptote at =0. As approaches positive infinit, approaches positive infinit. There is no vertical asmptote. 7. There is a vertical asmptote at =0. As approaches positive infinit, approaches positive infinit. There is no horizontal asmptote. 8. As approaches negative infinit there is a horizontal asmptote at =0. As approaches positive infinit there is a horizontal asmptote at =. There is no vertical asmptote. 9. There is a vertical asmptote at =0. As approaches positive infinit there is a horizontal asmptote at =0. As approaches negative infinit there is a horizontal asmptote at =. 0. There is a vertical asmptote at =. As approaches both positive and negative infinit there is a horizontal asmptote at =.. There is a vertical asmptote at =. As approaches both positive and negative infinit there is a horizontal asmptote at =.. Because when = 0, = which is undefined. 0. Because when =, = which is undefined. 0. = 5. = CK- PreCalculus Concepts

14 Chapter Functions and Graphs Answer Ke.0 Continuit and Discontinuit. This function is continuous.. This function is continuous.. This function is continuous.. This function is continuous. 5. Infinite discontinuit at = This function is continuous. 7. This function is continuous. 8. This function is continuous. 9. Removable discontinuit at =, infinite discontinuit at = 0, jump discontinuit at =. 0. Removable discontinuit at =.. Jump discontinuit at = 0... var, but should show f() has a jump discontinuit at =, a removable discontinuit at = 5, and another jump discontinuit at = 6.. var, but should show g() has a jump discontinuit at =, an infinite discontinuit at =, and another jump discontinuit at =.. var, but should show h() has a removable discontinuit at =, a jump discontinuit at =, and another jump discontinuit at = var, but should show j() has an infinite discontinuit at = 0, a removable discontinuit at =, and a jump discontinuit at =. CK- PreCalculus Concepts

15 Chapter Functions and Graphs Answer Ke. Function Composition. Here are the three graphs: f(): g(): h(): f(g()) = ( ). The negative values of the original parabola have been reflected across the -ais: - -. g(f()) = ( ) 5. The portion of the original parabola to the right of the -ais has been reflected across the -ais h(g()) = [( ) ] CK- PreCalculus Concepts 5

16 Chapter Functions and Graphs Answer Ke 7. The original parabola has been reflected across the -ais g(h()) = ( ) 9. The original parabola has been reflected across the -ais Here are the three graphs: j() = k() = m() = j(k()) = ( ). The graph looks the same as j(). CK- PreCalculus Concepts 6

17 Chapter Functions and Graphs Answer Ke k(m()) =. The graph looks the same as m(). 5. m(k()) = 6. The original square root graph is there, as well as its reflection across the -ais CK- PreCalculus Concepts 7

18 Chapter Functions and Graphs Answer Ke. Inverses of Functions. f() is shown in blue while f () is shown in red f () =. It is a function.. f(f ()) = ( ) =. f (f()) = ( ) =. g() is shown in blue while g () is shown in red g () = for g(g ()) = ( ) =. g (g()) = =. 7. h() is shown in blue while h () is shown in red. CK- PreCalculus Concepts 8

19 Chapter Functions and Graphs Answer Ke The inverse is = and is not a function. 9. You can see from the graph that the are inverses because the are smmetrical across the line =. 0. j() is shown in blue while j () is shown in red j () = +5. It is a function.. j(j ()) = ( +5 ) 5 = =. j (j()) = ( 5)+5 = =.. It is not: CK- PreCalculus Concepts 9

20 Chapter Functions and Graphs Answer Ke No. The inverse of g() is g () = e. 5. You could switch the and coordinates given in the original table to make the table for the inverse. CK- PreCalculus Concepts 0