2.2 notation and algebra of functions

1,033 views

Published on

Published in: Technology, Education

2.2 notation and algebra of functions

  1. 1. Notation and Algebra of Functions http://www.lahc.edu/math/precalculus/math_260a.html
  2. 2. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). Notation and Algebra of Functions
  3. 3. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Notation and Algebra of Functions
  4. 4. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Notation and Algebra of Functions
  5. 5. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions
  6. 6. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered),
  7. 7. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost).
  8. 8. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function.
  9. 9. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function. Its domain is D = {1, 2, 3, …}
  10. 10. Recall that the pizzas at Pizza Grande cost $8 each and there is $10 delivery charge so the cost is $50 for 5 pizzas (delivered). For x pizzas delivered, the cost is given by the expression “8x + 10”. Expressions are procedures for calculating future outputs and this leads us to the notion of “functions”. Definition: A function is a procedure that assigns each input x from a set D (the domain) to exactly one output y in a set R (the range). Notation and Algebra of Functions The pizza–cost expression “8x + 10” assigns each input x (the number of pizzas ordered), to a single output y (the final cost). So this procedure is a function. Its domain is D = {1, 2, 3, …} and its range is R = {various $-cost}.
  11. 11. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range.
  12. 12. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder.
  13. 13. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver.
  14. 14. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = the range R =
  15. 15. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R =
  16. 16. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}.
  17. 17. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}. b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder.
  18. 18. Notation and Algebra of Functions Example A. Is the following procedure a function? If it’s a function, describe its domain and its range. a. A procedure that takes any driver–license number as an input and outputs the name of the license holder. This is a function because each license number is assigned to exactly to one driver. The domain D = {all license numbers}, the range R = {names of all the people that are license holders}. b. A procedure that takes any driver–license number as an input and produces the name(s) of all of the cousins of that license holder. This is not a function because the license holder may have many cousins so there might be several outputs.
  19. 19. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions.
  20. 20. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y,
  21. 21. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  22. 22. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. the domain D the range R Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  23. 23. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range R Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  24. 24. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  25. 25. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  26. 26. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing:
  27. 27. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x
  28. 28. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = “Joe Blow”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x the output y
  29. 29. Notation and Algebra of Functions In mathematics, we often use the letters f, g, h,.. as names of functions. x y = f(x) f the domain D the range RIn applications, functions are named after their purposes. The function in example A might be named as the “Licen#-to-name” so if Joe Blow has the license number “123456”, then Licen#-to-name(123456) = “Joe Blow”. Suppose f is a function that assigns x to y from the domain D to the range R, we write that f(x) = y, and we summarize the setup with the following drawing: the input x the output y 123456 “Joe Blow” D ={Lic–numbers} R={Names} Licen#-to-name
  30. 30. Notation and Algebra of Functions The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.
  31. 31. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}.
  32. 32. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
  33. 33. We may define functions with tables, for example: x y=f(x) –1 4 2 3 5 –3 6 4 7 2 Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
  34. 34. We may define functions with tables, for example: x y=f(x) –1 4 2 3 5 –3 6 4 7 2 Notation and Algebra of Functions We search the table for outputs so f(2) = 3, f(6) = 4, etc.. There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function.
  35. 35. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
  36. 36. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Note that f(–1) = f(6) = 4, Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
  37. 37. x y=f(x) –1 4 2 3 5 –3 6 4 7 2 The domain of f is D = {–1, 2, 5, 6, 7}, and the its range is R = {4, 3, –3, 2}. Note that f(–1) = f(6) = 4, so that a function may assign multiple inputs to the same output. Notation and Algebra of Functions There are many ways to describe functions. The domain of the function “Licen#-to-name” is D = {all the license numbers}, its range R = {the names of all the license holders}. We may define functions by written instructions as in the case for the “Licen#-to-name” function. We may define functions with tables, for example: We search the table for outputs so f(2) = 3, f(6) = 4, etc..
  38. 38. Notation and Algebra of Functions Functions may be given graphically as the ones here:
  39. 39. For instance, Nominal Price(1975)  $0.50 Notation and Algebra of Functions (1975, $0.50) Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
  40. 40. Domain (Nominal Price) = {year 1918  2005} For instance, Nominal Price(1975)  $0.50 Notation and Algebra of Functions (1975, $0.50) 1918 2005 Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
  41. 41. Domain (Nominal Price) = {year 1918  2005} Range (Nominal Price) = {$0.20$2.51} For instance, Nominal Price(1975)  $0.50 Notation and Algebra of Functions (1975, $0.50) $0.20 $2.51 1918 2005 Nominal–price is the price that’s posted at the gas stations. Functions may be given graphically as the ones here:
  42. 42. Notation and Algebra of Functions (1975, $0.50) (1975, $1.85) Adjusted Price(1975)  $1.85 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
  43. 43. Notation and Algebra of Functions Domain (Adjusted Price) = {year 1918  2005} Adjusted Price(1975)  $1.85 (1975, $0.50) (1975, $1.85) 1918 2005 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
  44. 44. Notation and Algebra of Functions Domain (Adjusted Price) = {year 1918  2005} Range (Adjusted Price) = {$1.25$3.50} Adjusted Price(1975)  $1.85 (1975, $0.50) (1975, $1.85) 1918 2005 $1.25 $3.50 Adjusted–price is the inflation adjusted price in 2007–dollar. Functions may be given graphically as the ones here:
  45. 45. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f.
  46. 46. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically.
  47. 47. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out.
  48. 48. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. x y x y x y x yA. B. C. D.
  49. 49. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. Hence A and B below are graphs of functions x y x y x y x y A and B are functions because for each input x there is one output y. A. B. C. D. (x, y) (x, y)
  50. 50. Notation and Algebra of Functions Let f(x) = y be a function of numbers, the plot of all the points (x, f(x)) in the Cartesian system is the graph of the function f. The graph of a function can’t contain two or more points whose x–coordinates are the same i.e. having multiple points lining up vertically. Having multiple points with the same x–coordinate would violate the function rule of one–in–one–out. Hence A and B below are graphs of functions but C and D are not graphs of functions. x y x y (x, y2) y x y A and B are functions because for each input x there is one output y. A. B. C. D. C and D are not since there are multiple points lining up vertically. (x, y) (x, y) x (x, y1) (x, y3)
  51. 51. Notation and Algebra of Functions Most functions are given by mathematical formulas.
  52. 52. For example, f(X) = X2 – 2X + 3 = y Notation and Algebra of Functions Most functions are given by mathematical formulas.
  53. 53. For example, f(X) = X2 – 2X + 3 = y name of the function Notation and Algebra of Functions Most functions are given by mathematical formulas.
  54. 54. For example, f(X) = X2 – 2X + 3 = y name of the function Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
  55. 55. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
  56. 56. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box
  57. 57. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula.
  58. 58. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) =
  59. 59. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3
  60. 60. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y.
  61. 61. For example, f(X) = X2 – 2X + 3 = y name of actual formula the function The output Notation and Algebra of Functions Most functions are given by mathematical formulas. input box The input box holds the input for the formula. Hence f (2) means to replace x by (2) in the formula, so f(2) = (2)2 – 2(2) + 3 = 3 = y. Real functions are functions whose domain and range are subsets of real numbers (no complex numbers).
  62. 62. Domain of Functions
  63. 63. Domain of Functions There are two main restrictions to consider when determining the domains of real functions.
  64. 64. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0.
  65. 65. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6)
  66. 66. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0
  67. 67. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3
  68. 68. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}.
  69. 69. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}. b. f(x) =  2x + 6
  70. 70. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}. b. f(x) =  2x + 6 We can only extract square roots of nonnegative numbers
  71. 71. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}. b. f(x) =  2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0  x ≥ –3.
  72. 72. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}. b. f(x) =  2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0  x ≥ –3. So the domain = {all numbers x ≥ –3}.
  73. 73. Domain of Functions There are two main restrictions to consider when determining the domains of real functions. 1. The denominators can't be 0. 2. The radicands of even-order roots can't be negative. Example C. Find the domain of the following functions. a. f(x) = 1/(2x + 6) The denominator can’t be 0 i.e. 2x + 6 = 0  x = –3 So the domain = {all numbers except x = –3}. b. f(x) =  2x + 6 We can only extract square roots of nonnegative numbers hence 2x + 6 ≥ 0  x ≥ –3. So the domain = {all numbers x ≥ –3}. The requirement of having “the radicands ≥ 0” applies to the 4th root, or the 6th root, or any even-order root.
  74. 74. Domain of Functions c. Find the domain of f(x) = 1  x + 6 – 14
  75. 75. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 14
  76. 76. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 4
  77. 77. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. 4
  78. 78. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 4
  79. 79. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, – 5– 6 UDF 4
  80. 80. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, we get : – 5– 6 UDF – –+– – 4
  81. 81. Domain of Functions c. Find the domain of f(x) = The radicand of a 4th-root must be nonnegative. 1  x + 6 – 1 Hence x must satisfy the inequality 1 x + 6 – 1 > 0 This is a sign-charting problem. Put it in factored form: -x – 5 x + 6 > 0 Draw the real line and sample points, we get : – 5– 6 UDF – –+– – The domain consists of the non–negative portion i.e. the domain is the interval (–6 ,–5]. 4
  82. 82. Algebra of Functions We may form expressions using functions or use expressions as inputs for functions.
  83. 83. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) b. f(a+b)
  84. 84. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, b. f(a+b)
  85. 85. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, b. f(a+b)
  86. 86. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b)
  87. 87. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3
  88. 88. c. f(2a) – g(a + b) Algebra of Functions We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
  89. 89. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
  90. 90. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4 Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
  91. 91. c. f(2a) – g(a + b) = (2a)2 – 2*(2a) + 3 – [3(a + b) – 4] = 4a2 – 4a + 3 – 3a – 3b + 4 = 4a2 – 7a – 3b + 7 Algebra of Functions insert [ ] for subtraction We may form expressions using functions or use expressions as inputs for functions. Example D. Simplify if f(x) = x2 – 2x + 3, g(x) = 3x – 4. a. f(3) – g(4) f(3) = 32 – 2*3 + 3 = 6, g(4) = 3*4 – 4 = 8, so f(3) – g(4) = 6 – 8 = –2. b. f(a+b) = (a+b)2 – 2(a+b) + 3 = a2 + 2ab + b2 – 2a – 2b + 3
  92. 92. Notation and Algebra of Functions One important function-expression for any f(x) is its "difference quotient": h f(x+h) – f(x) where h is another variable.
  93. 93. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  94. 94. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  95. 95. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  96. 96. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  97. 97. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  98. 98. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  99. 99. Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h = h h(4x + 2h – 3) Example E. Let f(x) = 2x2 – 3x + 4, simplify h f(x+h) – f(x) One important function-expression for any f(x) is its "difference quotient":
  100. 100. Example E. Let f(x) = 2x2 – 3x + 4, simplify Notation and Algebra of Functions h f(x+h) – f(x) where h is another variable. h f(x+h) – f(x) h f(x+h) – f(x) = h 2(x+h)2 – 3(x+ h) + 4 – [2x2 – 3x + 4] = h 2x2 + 4xh + 2h2 – 3x – 3h + 4 – [2x2 – 3x + 4] = h 4xh + 2h2 – 3h = h h(4x + 2h – 3) = 4x + 2h – 3 One important function-expression for any f(x) is its "difference quotient":
  101. 101. Composition of Functions When one function is used as the input of another function, the outcome is called their composition.
  102. 102. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)).
  103. 103. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f.
  104. 104. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2)
  105. 105. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2))
  106. 106. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5
  107. 107. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5)
  108. 108. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20
  109. 109. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20
  110. 110. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2)
  111. 111. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2))
  112. 112. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1
  113. 113. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1
  114. 114. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x)
  115. 115. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3
  116. 116. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3 = –12x + 23
  117. 117. Composition of Functions When one function is used as the input of another function, the outcome is called their composition. Given f(x) and g(x), we define (f ○ g)(x) ≡ f (g(x)). In other words, input g(x) into f. Similarly, we define (g ○ f)(x) ≡ g (f(x)). Example F. Let f(x) = 3x – 5, g(x) = –4x + 3, simplify a. (f ○ g)(2) = f(g(2)) g(2) = –4*2 + 3 = –5 = f(–5) = 3(–5) – 5 = –20 b. (g ○ f)(2) = g(f(2)) f(2) = 3*2 – 5 = 1 = g(1) = –4*1 + 3 = –1 c. (g ○ f)(x) = g (f(x)) = –4(3x – 5) + 3 = –12x + 23 In general, (f○g)(x) = (g○f)(x).
  118. 118. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1)
  119. 119. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) b. (f○g○h)(1)
  120. 120. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) b. (f○g○h)(1)
  121. 121. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) b. (f○g○h)(1)
  122. 122. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 b. (f○g○h)(1)
  123. 123. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1))
  124. 124. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) (1, ½ )
  125. 125. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)), (1, ½ )
  126. 126. The Basic Language of Functions Exercise G. Given the functions f, g and h, estimate the outcomes of the following expressions. If it’s not defined, state so. x y = g(x) –1 4 2 3 5 –3 6 4 7 2 y = h(x) f(x) = 3x + 8 a. (h○g○f)(–1) b. (f○g○h)(1) (h○g○f)(–1) = h(g(f(–1)) = h(g(5)) = h(–3) ≈ 4 (f○g○h)(1) = f(g(h(1)) ≈ f(g(1/2)), this is UDF because x = ½ in not in the domain of g. (1, ½ )
  127. 127. Composition of Functions Equations may be posed in the language of functions.
  128. 128. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x)
  129. 129. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2
  130. 130. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 x2 – 4x – 5 = 0Setting one side 0:
  131. 131. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 x2 – 4x – 5 = 0Setting one side 0:
  132. 132. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0Setting one side 0:
  133. 133. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 Setting one side 0:
  134. 134. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 (x + 1)(x – 2) = 0 so x = –1, or 2, x2 – x – 2 = 0 Setting one side 0: Setting one side 0:
  135. 135. Composition of Functions Equations may be posed in the language of functions. Example H. Let f(x) = x2 – 4x – 3 and g(x) = –3x – 1, solve for x where a. f(x) = 2 b. f(x) = g(x) f(x) = 2 means that x2 – 4x – 3 = 2 (x + 1)(x – 5) = 0 so x = –1, or 5, or that f(–1) = f(5) = 2. x2 – 4x – 5 = 0 f(x) = g(x) means that x2 – 4x – 3 = –3x – 1 (x + 1)(x – 2) = 0 so x = –1, or 2, x2 – x – 2 = 0 i.e. f(–1) = g(–1) and f(2) = g(2) Setting one side 0: Setting one side 0:
  136. 136. Composition of Functions c. Solve for x where f(x) ≥ g(x).
  137. 137. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1.
  138. 138. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0
  139. 139. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0
  140. 140. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 2–1 (x + 1)(x – 2)
  141. 141. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
  142. 142. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
  143. 143. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2)
  144. 144. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2) The step of “setting one side 0” allows us to transform all equations and inequalities into problems of finding roots (zeros) or sign–charting.
  145. 145. Composition of Functions c. Solve for x where f(x) ≥ g(x). f(x) ≥ g(x) means that x2 – 4x – 3 ≥ –3x – 1. From (x + 1)(x – 2) = 0, we get two roots x = –1 or 2. Do the sign chart, testing x = 0 and the return is “–”. Setting one side 0 to transform it to a sign-charting problem: x2 – x – 2 ≥ 0 So f(x) ≥ g(x) for x’s from: (–∞, –1] U [2, ∞). Both roots have the order 1, hence both shoulder segments are + which are the regions that we want. 0 2–1 + + + – – – – + + + + (x + 1)(x – 2) The step of “setting one side 0” allows us to transform all equations and inequalities into problems of finding roots (zeros) or sign–charting. This step will be done as a routine from now on.

×