Lesson 1: Functions as ModelsA function happen to be a simple mathemical model or a piece of larger model.Functions describe situations where one quantity determines another. For ex-ample, the return on P10,000 invested at an annualized percentage rate of 4.25is a function of the length of time the money is invested. Because we continu-ally make theories about dependencies between quantities in nature and society,functions are important tools in the construction of mathematical models.In school mathematics, functions usually have numerical inputs and outputsand are often dened by an algebraic expression.
For example, the time inhours it takes for a car to drive 100 miles is a function of the cars speed in milesper hour, v; the rule T(v) = 100/v expresses this relationship algebraically anddenes a function whose name is T.The set of inputs to a function is called its domain. We often infer the do-main to be all inputs for which the expression dening a function has a value,or for which the function makes sense in a given context.A function can be described in various ways, such as by a graph (e.g.
, thetrace of a seismograph); by a verbal rule, as in, Ill give you a state, you give methe capital city; by an algebraic expression like f(x) = a + bx; or by a recursiverule. The graph of a function is often a useful way of visualizing the relation-ship of the function models, and manipulating a mathematical expression for afunction can throw light on the functions properties.Functions presented as expressions can model many important phenomena. Twoimportant families of functions characterized by laws of growth are linear func-tions, which grow at a constant rate, and exponential functions, which grow ata constant percent rate. Linear functions with a constant term of zero describeproportional relationships.
A graphing utility or a computer algebra system can be used to experimentwith properties of these functions and their graphs and to build computationalmodels of functions, including recursively dened functions.Example 1: The graph of a function h is shown below:(a) Find the values h(1) and h(5).(b) What is the domain and range of h?1Solution:(a) We see from the graph that the point (1,3) lies on the graph of h, so thevalue of h at 1 is h(1) = 3. While x = 5, the graph lies about 0.7 unit belowthe x- axis. Therefore, we estimate that h(5) = -0.7.
(b) Notice that h(x) is dened when 0 x 7, so the domain of f is theclosed interval 0,7. See that h takes on all values from the interval -2 to 4, sothe range of h is -2 y 4 = -2,4Example 2: Sketch the graph and nd the domain and range of the func-tions:(a) f(x) = 2x- 1(b) g(x) = x2Solution:(a)The equation of graph is y = 2x-1, and recognize this as being the equa-tion of a line with slope 2 and y-intercept -1.Recall: The slope- intercept form of the equation of a line y = mx + b.This enables us to sketch the graph below. The expression is dened for allreal numbers, so the domain of is the set of all real numbers, which we denoteby R. The graph shows that the range is also R.2(b) Since g(2) = 22= 4 and g(-1) = ( 1) 2= 1, we could plot the points (2,4)and (-1,1), together with a few other points on the graph, and join them to pro-duce the graph below.
The equation of the graph is y = x2, which represents aparabola. The domain of g is R. The range of g consists of all values of g(x),and that is the all numbers of the form x2. But x2 0 for all numbers x andany positive number of y is a square. Therefore, the range of g is y| y 0 =0, 1). This can also be seen from the gure below. Example 3: The cost of a pound of orange juice for three consecutuve weekis given by the table below:The Price of Orange Juice Week Week 1 Week 2 Week 3Cost 200 215 230What will be the cost of a pound of orange juice be in Week 4?Solution: The actual cost of a pound of orange juice in Week 4 will be de-termined by a number of factors, such as orange juice production, distribution,3sales, etc. These factors are the natural law governing orange juice cost.
Therecent cost of orange juice can be model as:x= the number of weeks since Week 1P(x) = the cost of a pound of orange juice at time x, in pesosThe table above have shown us the details: P(0) = 200, P(1) = 215, and P(2) =230. Then summarizing this information with the function will show us: P(x)= 200 + (15)x.Using the model, we can deduce that P(3) = 200 +(15)(3) = 245 pesos. We cannow predict that the cost of a pound of orange juice in Week 4 will be 245 pesos.
Note that this may or may not be accurate. The model between the relation-ship of orange juice and it’s cost is based entirely on an observation of previouspatterns.Example 4: The populations PB andPM of two colonies of penguins, alongthe Broughton and Montague Island of the New South Wales, respectively, havebeen successfully modeled by the following two functions:PB (t) = 2+(0.02)t3,P M (t) = (1:1) twhere the populations are measured in thousands of pairs of penguins and t ismeasured in years from the present. When are the two colonies of equal size?Solution: Since both of the functions are given by algebraic representationsformulas a natural instinct might be to try to solve the problem using an alge-braic method. In algebraic terms, the problem would be to solve the followingequation for t :2 + (0.
002)t3= (1 :1) t.However, this equation does not have an algebraic solution! (And it isnt becausewe just dont know how to solve it.) Algebraic methods of solution failed.
Does this mean that the problem has no solution?Before abandoning the algebraic approach entirely, it will serve as a clue, ifnot a complete solution. Notice that right now (t = 0)PB (0) = 2+(0.002)(0) 3= 2,P M (0) = (1:1) 0= 1.4The Broughton colony currently consists of 2000 penguins to the Montaguecolonys 1000. The only way the Broughton’s and Montague’s populations willever be equal at some point in the future is if the Montague colony outgrowsthe Broughton colony.
See that we must look at the long-term behavior of thefunctions. It is exactly this kind of big picture for which another method – thegraphical method – is best suited.Plot the two population functions and convert the algebraic representationsto graphical ones. Over the next 36 years, the plots are shown below: The top blue curve represents the Broughton colony. It appears that theBroughton colony not only has a head start (observed algebraically above, butnot so apparent here), but it is also rapidly outgrowing the Montague colony.
This would seem to settle things: The Montague colony will never equal thesize of the Broughton colony. Unfortunately, this is the wrong.While graphical methods are able to give the big picture of a functions behavior,there is always the question of how big is big enough. If we had looked at thepopulations for a period of time twice as long, we would have seen the following: In this view it is apparent that the Montague population does catch up withthe Broughton population, somewhere between 60 and 70 years from now. Wecould now use the zooming feature on our calculator or computer to nd the in-tersection point and get a better estimate of the time when the two populations5will be the same.Have we found all of the solutions? You should, at least, be skeptical by now.Perhaps a still larger viewing window would reveal that the Broughton popu-lation eventually retakes the Montague population. Extending the time scalefarther and farther into the future, however, shows no such trend.
(What doesit show?) After a bit of experimentation in this direction, we may be ready toconclude that the populations are equal only once. Again, we would be wrong.If we extend our models only a short time into the past, we see that the popu-lations were also equal a little less than ten years ago. The plots look like this: Example 5: New South Wales PenguinsSee previous exampleSolution: Notice that the algebraic problem of solving 2 + (0.002)t3= (1 :1) t(abandoned previously) is the same as solving 2 + (0.
002)t3- (1 :1) t= 0That is the same as nding the root of the function f(t) = 2 + (0.002) t3- (1 :1) t.Using the graphical observation that the populations appear to be equal at somepoint between t = 60 and t = 70, we might use the formula for f to calculateapproximate values of f(60) and f(70):First Iteration t 60 70f(t) 121.52 -101.75Since the value of the function changes sign between t = 60 and t = 70, we haveconrmed that a root lies at some point in-between. Suppose we look half wayin-between:6Second Iretationt 60 65 70f(t) 121.52 60.88 -101.
75Then we see that the value of f changes sign between t = 65 and t = 70, andtherefore the root must lie at some point in this interval. Computing the valuehalf way across this new interval leads to another table entry:Third Iteration t 60 65 67.5 70f(t) 121.52 60.88 -5.22 -101.75Then notice that the root is in the interval between t = 67.
5 and t = 70.The method is clear enough to keep subdividing the one interval on which fchanges sign and compute a new value for the table at the midpoint. Even-tually, after many iterations of this simple step, we will be able to computethe root to any degree of accuracy.
This may seem tedious, but its algorithmicnature makes it an ideal method for use on calculators or computers, which canperform tedious calculations very quickly.After ve more iterations, the root is bracketed between t = 67.27 and t =67.
34. We therefore, conclude that, to one decimal place, the root is at t =67.3. This much accuracy could have been achieved fairly quickly by graphingand zooming.
Unlike graphical methods, however, there is no limit to the degreeof accuracy that we may obtain by using the numerical method over and overagain.Conclusion: Good mathematical models use the strengths of one representa-tion to make up for the weaknesses of another. Good mathematical modelers,likewise, use the strengths of one method to make up for the anothers weak-nesses. Whether we are modeling a simple cause and eect relationship or acomplex physical system, we must look at problems from a variety of points ofview, and make use of all of the tools that are available to us.7Lesson 2: Evaluating FunctionsTo evaluate a function1.
Substitute the given value in the function of x.2.Replace all the variable xwith the value of the function.3.
Then compute and simplify the given function.Example 1: Given the function: f(x ) = 2 x+ 1, nd f(6).Substitute 6 in place holder x,f(6) = 2 x+ 1Replace all the variable of xwith 6,f(6) = 2(6) + 1Then compute function. f(6) = 12 + 1f (6) = 13Therefore, f(6) = 13. It can also write in ordered pair (6,13).Example 2: Given the function f(x ) = x2+ 2 x+ 4 when x= 4. Substitute-4 in the place holder x,f( 4) = x2+ 2 x+ 4Replace the all the variables with 6, f( 4) = ( 4) 2+ 2( 4) + 4f ( 4) = (16) + ( 8) + 4f ( 4) = 12Therefore, f( 4) = 12 or simply as ( 4;12) :Example 3: Given g(x ) = x2+ 2 x- 1. Find g(2y).
Answer in terms of y.g(2 y) = x2+ 2 x 1g (2 y) = (2 y)2+ 2(2 y) 1g (2 y) = 4 y2+ 4 y 1Therefore, 4( y)2+ 4 y 1:8Example 4: Givenf(x ) = 2 x2+ 4 x- 12, nd f(2 x+ 4).Solution:f(2 x+ 4) = 2 x2+ 4 x 12= 2(2 x+ 4) 2+ 4(2 x+ 4) 12= 2(2 x+ 4)(2 x+ 4) + 4(2 x+ 4) 12= 2(4 x2+ 16 x+ 16) + 4(2 x+ 4) 12= (8 x2+ 32 x+ 32) + (8 x+ 16) 12Combine like terms f(2 x+ 4) = 8 x2+ (32 x+ 8 x) + (32 + 16 12)= 8 x2+ 40 x+ 36= 2(2 x2+ 10 x+ 9)Therefore, f(2 x+ 4) = 2(2 x2+ 10 x+ 9).Example 5: Given f(x ) = x2-x – 4. If f(m ) = 8, compute the value of mSolution: Make the function f(x ) equivalent to f(m )x 2 x 4 = 8x 2 x 12 = 0( x 4)( x+ 3) = 0x 4 + 0; x+ 3 = 0x = 4; x= 3Therefore, the value of a can be either 4 or -3.9Exercises:Evaluate the functionsgiven:1.
p(x ) = 2x + 1, nd p(-2)2. p(x ) = 4 x, nd p(-4)3. g(n ) = 3 n2+ 6, nd g(8)4. g(x ) = x3+ 4 x, nd g(5)5. f(n ) = n3+ 3 n2, nd f(-5)6. w(a ) = a2+ 5 a, nd w(7)7.
p(a ) = a3- 4 a, nd p(-6)8. f(n ) = 4 3n+ 8 5, ndf(-1)9. f(x) = -1 + 1 4x;nd f(3 4)10. h(n) = n3+ 6 n, nd h(4)10Answers in Exercises:1. 52.
-163. 1984. 1455. -506. 847. -1928. 4 159. – 13 1610.
8811Lesson 3: Operations on FunctionsLet h(x) and g(x) be functions, and the operations on these two functions isshown below: Adding two functions as:(h+g)(x) = h(x)+g(x) Subtracting two functions as:(h-g)(x) = h(x) – g(x) Multiplying two functions as:(h g)(x) = h(x) g(c) Dividing two functions as:( h g)(x) = h(x ) g(x ) ; whereg(x ) 6= 0Example 1:Let f(x) = 4x + 5 and g(x) = 3x. Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g)(x). (f+g)(x) = (4x+5) + (3x) = 7x+5 (f-g)(x) = (4x+5) – (3x) = x+5 (f g)(x) = (4x+5) (3x) = 12 x2+5x (f g)(x) = 4x +5 3xExample 2:Let f(x)= 3x+2 and g(x)= 5x-1. Find (f+g)(x), (f-g)(x), (f g)(x), and ( f g)(x).
(f+g)(x) = (3x+2) + (5x-1) = 8x+1 (f-g)(x) = (3x+2) – (5x-1) = -2x+3 (f g) = (3x+2) (5x-1) = 15 x2+7x -2 (f g)(x) = 3x +2 5x 1Example 3:Let v(x) = x3and w(x) = 3 x2+5x. Find (v+w)(x), (v-w)(x), (v w)(x), and( v w)(x). (v+w)(x) = ( x3) + (3 x2+5x) = x3+ 3 x2+5x (v-w)(x) = ( x3) (3×2+5x) = x3 3x 2-5x (v w) = ( x3) (3×2+5x) = 3 x5+ 5 x4 (v w)(x) = ( x3 3x 2+5 x) = xx 2 x(3 x+5) = x2 3x +5Example 4:Let f(x) = 4 x3+ 2 x2+4x + 1 and g(x) = 3 x5+ 4 x2+8x-12. Find (f+g)(x),(f-g)(x), (f g)(x), and ( f g)(x).12(f+g)(x) = (4 x3+ 2 x2+4x+1) + (3 x5+ 4 x2+8x-12) = 3 x5+ 4 x3+ 6 x2+12x-11 (f-g)(x) = (4 x3+ 2 x2+4x+1) – (3 x5+ 4 x2+8x-12) = 3x 5+ 4 x3 2x 2-4x+13 (f g)(x) = (4 x3+ 2 x2+4x+1) (3 x5+ 4 x2+8x-12)= 12 x8+ 6 x7+ 12 x6+ 19 x5+ 40 x4 16×3+ 12 x2 40x 12 (f g)(x) = (4×3+2 x2+4 x+1) (3×5+4 x2+8 x 12)Example 5:Let h(x) = 1 and g(x) = x4 x3+ x2-1. Find (h+g)(x), (h-g)(x), (h g)(x),and ( h g)(x). (h+g)(x) = (1) + ( x4 x3+ x2-1) = x4 x3+ x2 (h-g)(x) = (1) – ( x4 x3+ x2-1) = x4 x3+ x2+2 (h g)(x) = (1) (x 4 x3+ x2-1) = x4 x3+ x2-1 (h g)(x) = 1 x4 x3+ x2 113Exercises:1.
If h(x) = 7x+3 and g(x) = 2 x2+1. Find (f+g)(x)2. If f(x) = x5-18 and g(x) = x2- 6x + 9, what is the vaue of (g-h)(x)?3. If t(x) = 25 x5and s(x) = 55 x8, what is the value of ( t s)(x)?4.
If v(x) = x3and w(x) = x2+ 4, solve (v w)(x)?5. If f(x) = 4x + 11 and g(x) = 5x + 9, nd (f+g)(x).6. If f(z) = 7z – 4 and g(z) = z-2, nd (f-g)(x).7. If f(x) =8 x2-20 and g(x) =-4, nd( f g)(x).
8. If f(x) = 2x+2 and g(x) = 9 x2, what is the value of (f g)(x)?9. If f(x) = 7 x2+ 8x -3 and g(x) = 7x, solve for (f g)(x)?10. If f(x) = 35 x8- 45x and g(x) = 5x, what is the value of ( f g)(x).14Answers to Operations on Functions Exercises:1.
2 x2+7x +42. x5 x2+ 6x – 273. 5x 11×34. x5+ 4 x35. 9x +206. 6z -27.
2x 2+ 58. 18 x3+ 18 x29. 49 x3+ 56 x2- 2110.7 x7- 915