1.2 Function Notation - Textbook [Questions + Answers]

CHECK Your Understanding 1. Evaluate, where f(x) = 23x. a) f(2) c) f(-4) e) f(a) b) f(0) f) f(36) 2. The graphs of y = f(x) and y = g(x) are shown. 4- 4 2 2 X -4-20 2.4 4-20 24 2 4 .. -2 -4 y = f(x) T -2 4 y = g(x) Using the graphs, evaluate a) f(1) b) g(-2) c) f(4)-g(2) d) x when f(x) = -3 the carton at 11:00 a.m. a) Use function notation to write an equation for this situation. b) How much will be left in the carton at 1:00 p.m.? c) At what time will 450 mL of milk be left in the carton? PRACTISING 4. Evaluate f(-1), f(3), and f(1.5) for a) f(x)=(x-2)2-1 b) f(x) = 2+3x-4x 1 5. For f(x) = 2x , determine a) f(-3) b) f(0) c) f(1)-f(3) 6. The graph of y = f(x) is shown at the right. a) State the domain and range of f. b) Evaluate. i) f(3) ii) f(5) iii) f(53) iv) f(5)-f(3) c) h(3-1) d) (2-5x) 7. For h(x) = 2x - 5, determine a) h(a) b) b(6+1) 8. Consider the function g(t) = 3t+5. a) Create a table of values and graph the function. b) Determine each value. i) g(0) iv) g(2)g(1) ii) g(3) v) g(1001)g(1000) iii) g(1) g(0) vi) g(a+1)g(a) 9. Consider the function f(s)=565+9. a) Create a table of values for the function. b) Determine each value. i) f(0) ii) f(1) iii) f(2) + iv) f(3) v) [f(2)-f(1)]-[f(1)-f(0)] vi) [(3)-f(2)]-[f(2)-(1)] c) In part (b), what do you notice about the answers to parts (v) and (vi)? Explain why this happens. 10. The graph at the right shows f(x) = 2(x-3)21. a) Evaluate f(-2). b) What does f(-2) represent on the graph of f? c) State the domain and range of the relation. d) How do you know that fis a function from its graph? 11. For g(x)45x, determine the input for x when the output of g(x) is a) -6 b) 2 c) 0 3-5 6 y = f(x) 2- 4-20 -21 2 4 6 8- y = 2(x-3)2-1- 4- 08 8 -4 -8- 12. A company rents cars for $50 per day plus $0.15/km. a) Express the daily rental cost as a function of the number of kilometres travelled. b) Determine the rental cost if you drive 472 km in one day. c) Determine how far you can drive in a day for $80. 13. As a mental arithmetic exercise, a teacher asked her students to think of a number, triple it, and subtract the resulting number from 24. Finally, they were asked to multiply the resulting difference by the number they first thought of. a) Use function noration to express the final answer in terms of the original number. b) Determine the result of choosing numbers 3, 5, and 10. c) Determine the maximum result possible. 14. The second span of the Bluewater Bridge in Sarnia, Ontario, is supported by two parabolic arches. Each arch is set in concrete foundations that are on opposite sides of the St. Clair River. The arches are 281 m apart. The top of each arch rises 71 m above the river. Write a function to model the arch. 15. a) Graph the function f(x) = 3(x-1)24. b) What does f(-1) represent on the graph? Indicate on the graph how you would find f(-1). c) Use the equation to determine i) f(2)-f(1) ii) 2/(3)-7 iii) f(1-x) 16. Let f(x) = x²+2x 15. Determine the values of x for which a) f(x) = 0 b) f(x)=-12 c) f(x) = -16 17. Let f(x) = 3x +1 and g(x)=2x. Determine values for a such that a) f(a) = g(a) b) f(a²) = g(2a) 18. Explain, with examples, what function notation is and how it relates to the graph of a function. Include a discussion of the advantages of using function notation. Extending 19. The highest and lowest marks awarded on an examination were 285 and 75. All the marks must be reduced so that the highest and lowest marks become 200 and 60. a) Determine a linear function that will convert 285 to 200 and 75 to 60. b) Use the function to determine the new marks that correspond to original marks of 95, 175, 215, and 255. 20. A function f(x) has these properties: • The domain of f is the set of natural numbers. • f(1) = 1 f(x + 1) = f(x) + 3x(x+1)+1 a) Determine f(2), f(3), f(4), f(5), and f(6). b) Describe the function. 1.2 Function Notation, pp. 22-24 1. a) f(2)=23(2) =2-6 =-4 b) f(0)2-3(0) =2-0 =2 c) f(-4)2-3(4) = 2-(-12) = 14 df(2)=2-3(2) 2- 1 2 3 2 e) f(a)23(a) = 2-3a f) f(3b)23(3b) =2-9b 2. a) From the graph, f(1) = 2 b) From the graph g(-2) = 4 c) From the graph, f(4) 1. g(2) cannot be read from the given graph; however, the graph appears to represent the function g(x)=x(x + 4) because it is a downward- facing parabola with zeroes at 0 and -4. Checking the value at x2 confirms this. Sog(2)-2(2+4) = -12, and f(4)g(2)-1-(-12) = 11 d) f(x)=-3 at x-3 and x-4 3. a) All values must have the same units, so 1.2 L must be converted to 1200 mL. Now, for x in minutes and f(x) in ml, the amount of milk in the carton at time x is f(x) = 1200-3x because the carton started with 1200 mL of milk and it loses 3 mL every minute. b) At 1:00 p.m., 2 hours will have passed. 2 hours 120 minutes, so the amount of milk left at 1:00 p.m. is f(120) 1 f(120) 1200-3(120) 1200-360 = 840 mL c) The amount of that must pass until there is 450 mL left is x such that 450-1200-3x 3x = 750 x 250 minutes 250 minutes 4 hours and 10 minutes. 4 hours and 10 minutes after 11:00 a.m. is 3:10 p.m. 4. a) f(-1) (-1-2)2-1 = (-3)2-1 =9-1 = 8 f(3)(32)2-1 = (1)2-1 1-1 = 0 f(1.5) (1.52)2-1 = (-0.5)2-1 = 0.25-1 = -0.75 b) f(-1)2+3(-1)-4(-1)2 -2-3-4 =-5 f(3)2+3(3)-4(3)2 2+9-36 = -25 f(1.5) 2+3(1.5)-4(1.5)² =2+4.5-9 =-2.5 5. a) f(-3) = f(-3)= 1 2(-3) -1 6 1 b) f(0) = 2(0) 1 0 Division by 0 is undefined, so the function is undefined at x = 0 c) f(1)-f(3) = = 1 1 2(1) 2(3) 1-6 11 26 1/2 1-3 1 1 ()+=+ d) f 4 = 1 2 1 + 2 1 3 =2 2 3 b) i) g(0) 3(0) + 5 =0+5 = 5 ii) g(3) 3(3) +5 =9+5 = 14 iii) g(1)g(0) 3(1)+5(3(0) + 5) 3+5-(0+5) =8-5 3 iv) g(2)g(1)3(2)+5(3(1) + 5) 6+5-(3+5) =11-8 = 3 v) g(1001)g(1000) 3(1001)+5(3(1000)+5) 30035 (3000 + 5) 30083005 = 3 vi) g(a+1)g(a) = 3(a+1)+5-(3(a) + 5) 3a+3+5 (3+5) 3a+8(3a+5) = 3 6. a) domain (-2, 2, 3, 5, 7}, range (1, 2, 3, 4, 5} b) i) From the graph, f(3) = 4 ii) From the graph f(5) = 2 iv) f(5)-f(3)=2-4 -2 7. a) h(a)=2a-5 iii) f(5-3)=f(2). From the graph, f(2) = 5 b) h(b+1)=2(b+1)-5 =2b+2-5 =2b-3 c) h(31)2(3-1)-5 =6c-2-5 = 6c-7 d) h(25x) = 2(2-5x)-5 410x5 -10x1 58 5 8 9. a) s(s) 0 9 1 4 2 1 3 0 b) i) f(0)=(0)2-6(0) +9 0-0+9 =9 ii) f(1) (1)6(1) +9 =1-6+9 =4 iii) (2) (2)2-6(2) +9 4-12+9 =1 iv) (3) (3)-6(3) +9 9-189 = 0 v) [f(2)=f(1)]-[f(1)-f(0)] (1-4)(49) = (-3)-(-5) = 2 = (-1)-(-3) = 2 vi) [f(3)-f(2)]-[f(2)-f(1)] (0-1)-(1-4) 8. a) 1g(1) ↑g(t) 20 0 1 15-g(t)=3t+5 2 11 10- 3 14 5 5 41701234 c) They are the same; they represent the second differences, which are constant for a quadratic function. 10. a) f(-2)2(-2-3)2-1 = 2(-5)2-1 = 2(25)-1 50-1 = 49 b) The y-coordinate of the point on the graph with x-coordinate -2 c) Any number can be substituted for x, so the domain is all real numbers. Since 2(x-3) is always positive, the minimum value for the range occurs when that quantity is 0, which will result in a value of 0 (x-2)(x+5) for the function. The function can take on any real value greater than -1. domain {x ∈ R), range {y ∈ Rly = -1} d) It passes the vertical line test 11.a)-64-5x 5x = 10 x=2 b) 245x 5x=2 2 5 x= c) 0 = 4-5x 5x = 4 4 x= d)=4-5x 5x = 17 5 17 25 x= 12. a) The company charges f(x)=0.15x + 50 because there is a fixed cost of $50, plus $0.15 for every kilometre travelled. b) The cost of travelling 472 km is f(472) 0.15(472) + 50 70.8+50 = $120.80 c) The distance travelled so that the cost is $80 will be x such that 800.15x + 50 0.15x = 30 x = 200 So $80 permits travelling up to 200 kilometres. 13. a) Let f(x) be the number that results from an input number of x. Then, f(x) = x(24-3x) b) f(3)3(24-3(3)) 3(24-9) 3(15) = 45 f(-5)-5(24-3(-5)) =-5(24-15) = -5(39) = -195 f(10)10(24-3(10)) = 10(24-30) 10(-6) -60 c) The maximum value for f(x) will occur at the value of x that is halfway between the two zeroes. Since f(x) = x(243x), the zeroes will occur when x = 0 or 24-3x0. 24-3x0 3x = 24 x=8 So the maximum value occurs at x = 4, and f(4)4(24-3(4)) = 4(24-12) = 4(12) = 48 14. Since the arch is a parabola, it will be described by a quadratic function. The first step is to locate the zeroes. The coordinate system can be chosen so that the left zero is at x = 0. From the statement that the other side of the arch is 281 metres away, so that zero will be at x281. So, the quadratic with these zeros is a "preliminary" function f(x)=ax(x281), where a is a constant that remains to be deter- mined. I knew that the top of the arch is 71 metres above the river. Since the maximum value of a parabola occurs at the value halfway between the zeroes, this means that f(140.5) 71. So 71a(140.5-281) 71a(140.5)(-140.5) 71a(-19740.25) a = -0.0036 The arch is described by the function f(x)=0.0036x(x-281) 15. a) f(x)=3(x-1)2-4 2 4 4 b) f(x) represents the y-coordinate of the point on the graph with x-coordinate-1. It is found on the graph by starting from 1 on x-axis, moving up to the curve, then moving across to the y-axis. c) i) f(2)-f(1) 3(2-1)-4-(3(1-1)2-4) 3(1)-4-(3(0)2-4) -3-4-4 =3 ii) 2f(3)72(3(3-1)2-4)-7 2(3(2)²-4)-7 2(3(4)-4)-7 2(124)-7 = 2(8)-7 16-7 = 9 iii) f(1-x)=3(1-x-1)2-4 = 3x²-4 16. a) Since f(x) = 0, 0x²+2x-15 0=(x+5)(x-3) = 3 or-5 b) Since f(x)=-12, -12x²+2x-15 0=x²+2x-3 0=(x+3)(x-1) 1 or-3 c) Since f(x) = -16, -16x²+2x-15 0=x²+2x+1 0=(x+1)(x + 1) =-1 or-1 In this case there is a duplication of roots, which occurs when the vertex of a parabola lies on the x-axis. 17. a) For f(a) = g(a), 3a+12-a 4a1 1 4 a= b) For f(a²) = g(2a), 3a²+12-20 3a²+2a7-0 -2V(-2)²-4(3)(-1) 2(3) -2±4 6 2-6 6 1 6 -or-1 a= = = 3 18. Answers may vary; for example: f(x) is defined as equal to an expression involving x, for each x-value in the functions domain; the graph of f(x) is the set of all points (x, f(x)) for which x is in the domain. 19. a) The points given are f(285)= 200 and f(75) 60. A linear function will have the form f(x) = mx + b, where m is the slope and b is the y-intercept. The slope can be found from the two points given: 200-60 285-75 140 210 2 23 3 b can be determined by substituting m and one of the points: 60 2 =(75) + b 6050+b b = 10 So the final equation is 2 f(x)=x +10 2 b) f(95) = (95) + 10 = 1 63+10 63 3 1 = 73- 3 2, f(175) = (175) + 10 NIM NIO 116+10 116 = 126 771- =