1.2 Function Notation - Textbook [Questions + Answers]

CHECK Your Understanding1. Evaluate, where f(x) = 2-3x.a) f(2)c) f(-4)e) f(a)b) f(0)d) 1(1)f) f(36)2. The graphs of y f (x) = and y=g(x) = are shown.yy4422X-4 -202-2024-2-2y=f(x)y=g(x)-4-4Using the graphs, evaluatea) f(1)c) f(4)-g(2) -b) 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?PRACTISING4. Evaluate f(-1),f(3),andf(1.5) fora) f(x) = (x-2)2-1 b) f(x) )=2+3x-4x25. Fora) f(-3)b) f(0)c) f(1) - f(3)d) A(1)+1(s)y6. The graph of y = f(x) is shown at the right.a) State the domain and range of f.6y=f(x)b) Evaluate.4i) f(3)iii) f(5 - 3).2ii) f(5)iv) f(5) - f(3)x7. For b(x) = 2x - 5, determine6a) b(a)c) b(3c - 1)-2b) b(b + 1)b(2-5x)8. Consider the function g(t) = 3++5.a) Create a table of values and graph the function.b) Determine each value.i) &(0)iv) g(2) - g(1)ii) g(3)v) g(1001) - g(1000)iii) g(1) - &(0)vi) g(a+1)- - g(a)9. Consider the function f(s) = s2 - 6s 9.a) Create a table of values for the function.b) Determine each value.i) f(0)iv) f(3)ii) f(1)v) [f(2)-f(1)]-[f(1)-f(0)] - - -iii) f(2)vi) [f(3)-f(2)]-[f(2)-f(1)] - -c) In part (b), what do you notice about the answers to parts (v) and (vi)?ymExplain why this happens.10. The graph at the right shows f(x) = 2(x - 3)2-1.y=2(x-3)2-14Ka) Evaluate f(-2).b) What does f(-2) represent on the graph of f?8c) State the domain and range of the relation.-4d) How do you know that fis a function from its graph?-811. For g(x) = 4 - 5x, determine the input for x when the output of g(x) isa)-6b) 2c)0d) 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 kilometrestravelled.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 aAnumber, triple it, and subtract the resulting number from 24. Finally, they wereasked to multiply the resulting difference by the number they first thought of.a) Use function notation to express the final answer in terms of the originalnumber.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 byTtwo parabolic arches. Each arch is set in concrete foundations that are onopposite sides of the St. Clair River. The arches are 281 m apart. The top ofeach arch rises 71 m above the river. Write a function to model the arch.15. a) Graph the function f(x) = 3(x - 1)2 - 4.b) What does f(-1) represent on the graph? Indicate on the graph howyou would find f(-1).c) Use the equation to determinei) f(2) - f(1) ii) 2f(3) - 7 iii) f(1 - x)16. Let f(x) = x2 + 2x - 15. Determine the values of x for whicha) f(x) = 0b) f(x) = -12c) f(x) = -1617. et (f)x = 3x + 1 and g(x) = 2 - X. Determine values for a such thata) f(a) = g(a)b) f(a2) = g(2a)18. Explain, with examples, what function notation is and how it relates to theCgraph of a function. Include a discussion of the advantages of using functionnotation.Extending19. 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 become200 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 originalmarks 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) = 1f(x+1) = f(x) + 3x(x + 1) + 1a) Determine f(2), f(3),f(4),f(5) and f(6).b) Describe the function. 1.2 Function Notation, pp. 22-241.a) f(2) = 2 - 3(2)6b) f(0) = 2 - 3(0)= 2 0= 2c) f(-4) = 2 3(4)= 2 (-12)=142e) f(a) = 2 - 3(a)2 - 3af) f(3b) - 2 - 3(3b)2 - 9b2. a) From the graph, f(1) = 2b) From the graph g(-2) = 4c) From the graph, f(4) = --- 1. g(2) cannot beread from the given graph; however, the graphappears 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 x = -2 confirms this.So g(2) = -2(2+4)= -12, andf(4) - g(2) = -1 - (-12)= 11d) f(x) = - 3 at x = - 3 and x = - 43. a) All values must have the same units, SQ1.2L must be converted to 1200 mL. Now, for xin minutes and f(x) in mL, the amount of milkin the carton at time x is f(x) = 1200 - 3xbecause the carton started with 1200 mL ofmilk and it loses 3 mL every minute.b) At 1:00 p.m., 2 hours will have passed.2hours = 120 minutes, so the amount of milkleft at 1:00 p.m. is f(120)nf(120) = 1200 - 3(120)= 1200 - 360= 840mLc) The amount of that must pass until there is450 mL left is x such that450 = 1200 - 3x3x - 750x = 250 minutes250 minutes = 4 hours and 10 minutes. 4 hoursand 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= 8f(3) = (3 - 2)2 - 1= (1)2 - 1= 1 1= 0f(1.5) = (1.5 - 2)2 - 1= (-0.5)2 - 1= 0.25 1= -0.75b) f(-1) = 2 + 3(-1)-4(-1)2= 2 3 - 4= -5f(3) = 2 + 3(3)-4(3)29 36= -25f(1.5) 2 + 3(1.5)-4(1.5)2:4.5 -= -2.5 b)i) ((g(0) = 3(0)= 0+5=5f(-3)ii) g(3) = 3(3) + 51= 9+5= 141iii) g(1) - g(0) = 3(1) + 5 (3(0) + 5)0= 3+5 (0+5)= 8-5Division by 0 is undefined, so the function is= 3undefined at x = 01iv) (2) - g(1) = 3(2) 5 - (3(1) + 5)f(1)-f(3)= 6+5 (3+5)2(3)= 11 81= 32g(1001) - g(1000)1= 3(1001) + 5 - (3(1000) + 5)3= 3003 + 5 - (3000 + 5)= 3008 - 30053vi) g(a+1) - g(a)11= 3(a+1)+5- - (3(a) + 5)= 3a+3+5 - (3a+5)3= + 8 - (3a +5)= 39.a)2-Sf(s)3096. a) domain {-2,2,3,5,7},14range {1,2,3,4,5}21b) i) From the graph, f(3) = 430ii) From the graph f(5) = 2iii) - 3) = f(2). From the graph, f(2) = 5b) f(0) =0-0+9iv) f(5) - f(3) = 2 - 4= 9= -27. a) h(a) = 2a - 5ii) f(1) = (1)2 6(1) +9= -6+9b) h(b + 1) = 2(b 1 ) - 54= 2b + 5= 2b 3iii) f(2) = (2)2 6(2) +9= 12+9c) h (3c - 1) = 2(3c - 1) - 5= 1= 6c 5= 6c 7iv) f(3) = (3)2 6(3) + 9= 9 18d) h(2 - 5x) = 2(2 - 5x) - 5= 0= 4 10x 5= 10x - 1v) [f(2) = f(1)] - [f(1) - f(0)]a)= (1 4) (4-9) -tg(t)g(t)0520-= (-3)-(-5)15g(t)=3t+5= 21821110vi) [f(3) - f(2)] - [f(2) - f(1)]= (0 - 1)- (1 - 4)3145= (-1)- (-3)4172520 c) They are the same; they represent the second13. a) Let f(x) be the number that results fromdifferences, which are constant for a quadratican input number of Then, f(x) = x(24 - 3x)function.b) f(3) = 3(24 - 3(3))10. f(-2) = 2(-2 3)2 1= 3(24-9)= 2(-5)2 1= 3(15)= 2(25) 1= 45= 50 1f(-5) = -5(24 - 3(-5))= 49= -5(24 - 15)b) The y-coordinate of the point on the graph= -5(39)with x-coordinate -2= -195c) Any number can be substituted for x, so thef(10) = 10(24 - 3(10))domain is all real numbers. Since 2(x-3)2is -= 10(24 - 30)always positive, the minimum value for the= 10(-6)range occurs when that quantity is 0, which will= -60result in a value of 0 = (x - (x + 5) for thec) The maximum value for f(x) will occur atfunction. The function can take on any realthe value of x that is halfway between the twovalue greater than -1.zeroes. Since f(x) = x(24 - 3x), the zeroesdomain {x E R}, range [yERlyz-1}will occur when x = 0 or 24 - 3x = 0.d) It passes the vertical line test24 - 3x = 011. a) -6 = 4 - 5x3x = 245x = 10x 8x 2So the maximum value occurs at x = 4, andb) = 4 - 5xf(4) = 4(24 - 3(4))5x 2= 4(24 - 12)2= 4(12)5= 484 5x14. Since the arch is a parabola, it will bedescribed by a quadratic function. The first step4is to locate the zeroes. The coordinate system5can be chosen so that the left zero is at x = 0.From the statement that the other side of thearch is 281 metres away, SO that zero will be atx = 281. So, the quadratic with these zeros is a"preliminary" function f(x) = ax(x - 281),where a is a constant that remains to be deter-17mined. I knew that the top of the arch is 7125metres above the river. Since the maximum12. a) The company chargesvalue of a parabola occurs at the value halfwayf(x) = 0.15x + 50 because there is a fixed costbetween the zeroes, this means thatof $50, plus $0.15 for every kilometre travelled.f(140.5) = 71. Sob) The cost of travelling 472 km is71 = a(140.5 - 281)f(472) = 0.15(472) + 5071 = a(140.5) (-140.5)= 70.8 + 5071 = a(-19740.25)= $120.80a = -0.0036c) The distance travelled so that the cost is $80The arch is described by the functionwill be x such thatf(x) = -0.0036x(x - 281)80 - 0.15x + 500.15x = 30x = 200So $80 permits travelling up to 200 kilometres. 15. a)=3(x-1)2-4 -b) For f(a2) = g(2a),3a2+1=2-2a4-2V(-2)2-4(3)(-1)a22(3)-4-20244b) f(x) represents the y-coordinate of the pointon the graph with x-coordinate - -1. It is found onorthe graph by starting from - -1 on x-axis, moving18. Answers may vary; for example: f(x) isup to the curve, then moving across to the y-axis.defined as equal to an expression involving x,c) f(2)-f(1)for each x-value in the functions domain; the= 3(2-1)2-4-(3(1-1)2- - - 4)graph of f(x) is the set of all points (x,f(x))= 3(1)2 4 - (3(0)2for which x is in the domain.= 3 4 419. a) The points given are f(285) = 200 and3f(75) = 60. A linear function will have theii) 2f(3) - 7 = 2(3(3 - 1)2 - 4) - 7form f(x) = mx + b, where m is the slope and2(3(2)2 4) - 7b is the y-intercept. The slope can be found2(3(4) 4) - 7from the two points given:2(12 4) - 7= 2(8) 7= 16 79iii) f(1 x) = 3(1 4= 3x2 416. a) Since f(x) = 0,30=x+2x-15b can be determined by substituting m and= (x+5)(x- - 3)one of the points:3 or -5b) Since f(x) = -12,x2 + 2x - 1560 50 + b==++2x-3b 100 = (x+3)(x - 1)So the final equation is= 1 or -310c) Since f(x) = -16,x2 2x - 15+0 = x + 2x + 10 = (x+1)(x+ 1)= -1or-1In this case there is a duplication of roots,which occurs when the vertex of a parabola lies7312on the x-axis.17. a) For f(a) = g(a),f(175)=~(175) +3a+1=2 - a4a = 1101a4=12623 f(215)=3(215)+10+10= 1431/1 10=f(255) 10170 + 10= 18020. a) f(2) = f(1) + 3(1) (1+1) + 1= + 3(2) + 1= + 6+1= 8f(3) = f(2) + 3(2) (2 + 1) + 1= 8 + 6(3) + 1= + 18 + 1= 27f(4) = f(3) + 3(3) (3 + 1) + 1= 27 + 9(4) + 1= 27 + 36 + 1= 64f(5) = f(4) + 3(4) (4 + 1) + 1= 64 + 12(5) + 1= 64 + 60 + 1= 125f(6) = f(5) + 3(5)( (5 + 1) + 1= 8 + 15(6) + 1= 125 + 90 + 1= 216b) For each X, it appears that f(x) = x3. Thiscan be verified by noting that if f(x) = x3,f(x + 1) = (x+1)3 ==x3+3x2+3x + 1= f(x)+3x(x+1) + 1as was stated in the problem.