1. Functions Prepared by: Richard Mitchell Humber College 4

2. CASE STUDY

3. 4.1 – FUNCTIONS AND RELATIONS

4. Relations f(x) = 3x – 5 (Functional Notation) y = 3x 3 (Power) y = 3x 2 + 4 (Quadratic) y = 3 sin 2x (Trigonometric ) y = 5 2x (Exponential) y = log(x + 2) (Logarithmic) Relations 4.1-DEFINITIONS-Pages 111 to 112 Each value of x results in only one value of y. In these relations, y is a function of x. Each value of x results in two values of y. In these relations, y is not a function of x. THESE ARE NOT FUNCTIONS THESE ARE FUNCTIONS

5. 4.1-EXAMPLE 3-Page 112 Relations Domain (All of the x values) Range (All of the y values) THIS IS NOT A FUNCTION Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are two points associated with the same x value: (2, -3) and (2, 3). This one value, x = 2, gives two possible y values.

6. 4.1-EXAMPLE 4-Page 113 Relations Domain (All of the x values) Range (All of the y values ) THIS IS A FUNCTION Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are no duplicates of x values in the data set so this relation is a function.

7. 4.1-DEFINITIONS-Page 114 Graphing Domain (All of the x values) Range (All of the y values) THIS IS A FUNCTION Members of the Domain cannot repeat (only the y values can repeat). There can only be one y value associated with every x. Here, there are no duplicates of x values in the data set so this relation is a function. x-values (Domain) y-values (Range) 00 +13 +26 +39 +412 +515 0 0 1 3 2 6 3 9 4 12 515

8. 4.1-EXAMPLE 5-Page 116 Vertical Line Test THIS IS A FUNCTION A vertical line drawn anywhere on the curve intersects only at one point therefore it is a function. THIS IS NOT A FUNCTION A vertical line drawn anywhere on the curve intersects at two points therefore it is not a function.

9. 4.1-DEFINITIONS-Pages 117 to 118 The equation is a function rule that associates exactly one y with one x. (Domain x and Range y are Real Numbers). The ordered pair is a function since each value of x (load) is associated with one value of y (stretch). (Domain x and Range y are Real Numbers). Load (Kg) Domain 012345678 Stretch (cm) Range 00.50.91.42.12.43.03.64.0 The table of ordered pairs is a function since each value of x (load) is associated with one value of y (stretch). (Domain x and Range y are Real Numbers). I. Equations 6y + 2x = 6 Types of Functions II. Ordered Pairs A load of 6 kg causes a spring to stretch 3 cm. The ordered pair that results is (6, 3) where load is first and distance is second. III. Table of Values IV. Verbal Statement Write an equation for the volume of a cone in terms of its base (75 units) and altitude. V = 1/3 x (base area) x (altitude) V = 1/3 x (75) x H V = 25 H The formula is a function since each value of H (altitude) is associated with one value of V (volume). (Domain H and Range V are Real Numbers). V. Graphs f(x) = x 2 - 4x - 3 The graph is a function since each value on the x (axis) is associated with one value on the y (axis). (Domain x and Range y are Real Numbers).

10. 4.1-DEFINITIONS-Pages 118 to 120 Test what values of x and y do not work. Domain (x) and Range (y) will be those values that do work. Domain (x) and Range (y) E Example 15 Example 16 Example 18 Every value of x gives a real value of y. No value of x will make y negative. Test what values of x and y do not work. Domain (x) and Range (y) will be those values that do work. Any value of x less than 2 will make the quantity under the radical sign negative. An x value of 2 gives a y value of zero. An x value larger than 2 gives a y value greater than zero. An x value of 4 will make the quantity in the denominator equal to zero. Any value of x greater than 4 will make the quantity under the radical sign negative. All values of y will be positive.

11. 4.2 – FUNCTIONAL NOTATION

12. Explicit Form (One variable is isolated on one side) y = 2x 3 + 5 z = ay + b x = 3z 2 + 2z – 5 Implicit Form (One variable is not isolated to one side) y = x 2 + 4y x 2 + y 2 = 25 w + x = y + z + x Dependent and Independent Variables (Value of Dependent Variable y depends upon value of Independent Variable x) y = x + 5y = 2x 2 – 3 4.2-DEFINITIONS-Page 121

13. Manipulating Functions (Re-arranging) If 2x + y = 5, then y = f(x) x = f(y) f(x,y) = 0 y = 5 – 2x x = 2x + y – 5 = 0 If y – 4x = 5 – z, then y = f(x,z) x = f(y,z) z = f(x,y) f(x,y,z) = 0 y = 4x – z + 5 x = z = 4x – y + 5 4x-y-z+5 = 0

14. Write the equation y = 2x – 3 in the form x = f(y) x = f(y) 2x = y + 3 x = 4.2-EXAMPLE 27-Page 123

15. Write the equation y = 3x 2 – 2x in the form f(x,y) = 0 f(x,y) = 0 3x 2 – 2x – y = 0 ANS: 3x 2 – 2x – y = 0 4.2-EXAMPLE 29-Page 123

16. Substitution into Functions Given f(x) = (x) 3 – 5(x), find f(2) f(2) = (2) 3 – 5(2) f(2) = 8 – 10 f(2) = -2 ANS: f(2) = -2 4.2-EXAMPLE 30-Page 123 f(x) = (x) 3 – 5(x)

17. Given y(x) = 3(x) 2 – 2(x), find y(5) y(x) = 3(x) 2 – 2(x) y(5) = 3(5) 2 – 2(5) y(5) = 3(25) – 10 y(5) = 75 – 10 y(5) = 65 ANS: y(5) = 65 4.2-EXAMPLE 31-Page 124

18. Given f(x) = (x) 2 – 3(x) + 4, find f(x) = (x) 2 – 3(x) + 4 f(x) = (x) 2 – 3(x) + 4 f(x) = (x) 2 – 3(x) + 4 f(5) = (5) 2 – 3(5) + 4 f(2) = (2) 2 – 3(2) + 4 f(3) = (3) 2 – 3(3) + 4 f(5) = 25 – 15 + 4 f(2) = 4 – 6 + 4 f(3) = 9 – 9 + 4 f(5) = 10 + 4 f(2) = -2 + 4 f(3) = 0 + 4 f(5) = 14 f(2) = 2 f(3) = 4 ( con’t ) f(5) – 3∙f(2) 2∙f(3) 4.2-EXAMPLE 32-Page 124

19. Given f(x) = (x) 2 – 3(x) + 4, find (where f(5) = 14 f(2) = 2 and f(3) = 4) f(5) – 3∙f(2) 2∙f(3) = (14) – 3∙(2) 2∙(4) = 14 – 6 8 = 8 8 f(5) – 3∙f(2) 2∙f(3) ANS: 1 4.2-EXAMPLE 32-Page 124

20. Given f(x) = 3(x) 2 – 2(x) + 3, find f(5a) f(x) = 3(x) 2 – 2(x) + 3 f(5a) = 3(5a) 2 – 2(5a) + 3 f(5a) = 3(25a 2 ) – 10a + 3 f(5a) = 75a 2 – 10a + 3 ANS: f(5a) = 75a 2 – 10a +3 4.2-EXAMPLE 33-Page 124

21. Given f(x) = 5(x) – 2, find f(x + a) f(x) = 5(x) – 2 f(x + a) = 5(x + a) – 2 f(x + a) = 5x + 5a – 2 ANS: f(x+ a) = 5x + 5a – 2 4.2-EXAMPLE 36-Page 125

22. 2∙g(3) + 4∙h(9) f(5) (con’t) 4.2-EXAMPLE 38-Page 125

23. = 2∙(9) + 4∙(3) (15) = 18 + 12 15 = 30 15 ANS: 2 4.2-EXAMPLE 38-Page 125 2∙g(3) + 4∙h(9) f(5)

24. Given f(x,y,z) = 2(y) – 3(z) + x, find f(3,1,2) f(3,1,2) = 2(1) – 3(2) + 3 f(3,1,2) = 2 – 6 + 3 f(3,1,2) = -4 + 3 ANS: f(3,1,2) = -1 4.2-EXAMPLE 39-Page 125

25. 4.3 – COMPOSITE FUNCTIONS AND INVERSE FUNCTIONS

26. Given g(x) = (x) + 1, find g(2), g(z 2 ) and g[f(x)] g(x) = (x) + 1 g(x) = (x) + 1 g(x) = (x) + 1 g(2) = (2) + 1 g(z 2 ) = (z 2 ) + 1 g[f(x)] = (f(x)) + 1 g(2) = 3 g(z 2 ) = z 2 + 1 g[f(x)] = f(x) + 1 ANS: g(2) = 3 ANS: g(z 2 ) = z 2 + 1 ANS: g[f(x)] = f(x) + 1 4.3-EXAMPLE 40 (a), (b) and (c)-Page 128

27. Given g(x) = (x) 2 and f(x) = (x) + 1, find the following g[f(x)] f[g(2)] g[f(2)] g(x) = (x) 2 f(x) = (x) + 1 g(x) = (x) 2 g[f(x)] = (f(x)) 2 f[g(2)] = (g(2)) + 1 g[f(2)] = (f(2)) 2 = (x + 1) 2 = (2 2 ) + 1 = (2 + 1) 2 = x 2 + 2x + 1 = 5 = 9 ANS: x 2 + 2x + 1 ANS: 5 ANS: 9 4.3-EXAMPLE 42 (b), (c) and (d)-Page 129

28. Find the inverse f -1 (x) of the function y = f(x) = (x) 3 y = f(x) = (x) 3 4.3-EXAMPLE 45-Page 130 Step 1: Solve the given equation for x. Step 2: Interchange the x and y variables.

29. Find the inverse f -1 (x) of the function y = f(x) = 2x + 5 y = f(x) = 2x + 5 4.3-EXAMPLE 46-Page 130 Step 1: Solve the given equation for x. Step 2: Interchange the x and y variables.

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