Figure 1 gives a graphical representation of the number of male and female students regis- tered at a certain university in the period 1986 to 1997.
f(t)
m(t)
86 97
86 97 t
t
m(t) ⫹ f(t)
m(t)
86 91 97
60 100
20
Number in thousands
Male Female
Year Figure 1
Letf (t )andm(t )denote the number of female and male students in yeart, whilen(t ) denotes the total number of students. Of course,
n(t )=f (t )+m(t )
S E C T I O N 5 . 2 / N E W F U N C T I O N S F R O M O L D 133 The graph of the total numbern(t )is obtained by piling the graph off (t )on top of the graph ofm(t ). Suppose in general thatf andgare functions which both are defined in a setA of real numbers. The functionF defined by the formulaF (x)=f (x)+g(x)is called the sumoff andg, and we writeF =f+g. The functionGdefined byG(x)=f (x)−g(x) is called thedifferencebetweenf andg, and we writeG=f −g.
Sums and differences of functions are often seen in economic models. Consider the following typical examples.
E X A M P L E 1 The cost of producingQunits of a commodity isC(Q). The cost per unit of output, A(Q)=C(Q)/Q, is called the average cost.
A(Q)=C(Q)/Q (average cost)
If, in particular,C(Q)=aQ3+bQ2+cQ+dis a cubic cost function of the type shown in Fig. 4.7.2, the average cost is
A(Q)=aQ2+bQ+c+d/Q, Q >0
ThusA(Q)is a sum of a quadratic functiony=aQ2+bQ+cand the hyperbolay =d/Q.
Figure 2 shows how the graph of the average cost functionA(Q)is obtained by piling the graph of the hyperbolay =d/Qonto the graph of the parabolay=aQ2+bQ+c.
Note that for small values ofQthe graph ofA(Q)is close to the graph ofy =d/Q, while for large values ofQ, the graph is close to the parabola (sinced/Qis small whenQ is large).
aQ2⫹ bQ ⫹ c y
Q
d兾Q y
Q
d兾Q aQ2⫹ bQ ⫹ c A(Q) y
Q Figure 2 A(Q)=(aQ2+bQ+c)+d/Q=C(Q)/Q
LetR(Q)denote therevenuesobtained by producing (and selling)Qunits. Then theprofit π(Q)is given by
π(Q)=R(Q)−C(Q)
An example showing how to construct the graph of the profit functionπ(Q)is given in Fig. 3. In this case the firm gets a fixed pricepper unit, so that the graph ofR(Q)is a straight line through the origin. The graph of−C(Q)must be added to that ofR(Q). The production level which maximizes profit isQ∗.
Q*
C(Q) R(Q)
π(Q) y
Q
Figure 3 π(Q)=R(Q)−C(Q)
Products and Quotients
Iff andgare defined in a setA, the functionF defined byF (x)=f (x)ãg(x)is called the productoff andg, and we putF =fãg(orf g). The functionFdefined whereg(x)=0 byF (x)=f (x)/g(x)is called thequotientoff andg, and we writeF =f/g. We have already seen examples of these operations. It is difficult to give useful general rules about the behaviour of the graphs off gandf/ggiven the graphs off andg.
Composite Functions
Suppose the demand for a commodity is a functionxof its pricep. Suppose that pricepis not constant, but depends on timet. Then it is natural to regardxas a function oft.
In general, ify is a function ofu, anduis a function ofx, theny can be regarded as a function ofx. We callyacomposite functionofx. If we denote the two functions involved byf andg, withy =f (u)andu=g(x), then we can replaceubyg(x)and so writeyin the form
y=f g(x) Note that when computingf
g(x)
, we first applygtoxto obtaing(x), and then we apply f tog(x). The operation of first applyinggtox and thenf tog(x)defines acomposite function. Hereg(x)is called thekernel, orinterior function, whilefis called theexterior function.
NOTE 1 The function that mapsx tof g(x)
is often denoted byf g. This is read as
“f ofg” or “f afterg”, and is called thecompositionoff withg. Correspondingly,gf denotes the function that mapsxtog
f (x)
. Thus, we have (fg)(x)=f
g(x)
and (gf )(x)=g f (x)
Usually,f gandgf are quite different functions. For instance, ifg(x)=2−x2 and f (u)=u3, then(fg)(x)=(2−x2)3, whereas(gf )(x)=2−(x3)2 =2−x6; the two resulting polynomials are not the same.
S E C T I O N 5 . 2 / N E W F U N C T I O N S F R O M O L D 135 It is easy to confusefgwithfãg, especially typographically. But these two functions are defined in entirely different ways. When we evaluate f g atx, we first compute g(x)and then evaluatef atg(x). On the other hand, the productf ãgoff andgis the function whose value at a particular numberxis simply the product off (x)andg(x), so (fãg)(x)=f (x)ãg(x).
Many calculators have several built-in functions. When we enter a numberx0 and press the key for the functionf, we obtainf (x0). When we compute a composite function givenfandg, and try to obtain the value off
g(x)
, we proceed in a similar manner: enter the numberx0, then press thegkey to getg(x0), and again press thef key to getf (g(x0)). Suppose the machine has the functions 1/x and √x . If we enter the number 9, then press 1/x followed by √x ,we get 1/3 =0.33. . .The computation we have performed can be illustrated as follows:
1/x √
x
9 −→ 1/9 −→ 1/3.
Using function notation,f (x)=√
xandg(x)=1/x, sof (g(x))=f (1/x)=√ 1/x = 1/√
x. In particular,f (g(9))=1/√
9=1/3.
E X A M P L E 2 Write the following as composite functions:
(a) y=(x3+x2)50 (b)y =e−(x−μ)2 (μis a constant) Solution:
(a) You should ask yourself: What is the natural way of computing the values of this function? Given a value of x, you first computex3 +x2, which gives the interior function,g(x) = x3 +x2. Then take the fiftieth power of the result, so the exterior function isf (u)=u50. Hence,
f (g(x))=f (x3+x2)=(x3+x2)50
(b) We can choose the interior function asg(x)= −(x−μ)2and the exterior function as f (u)=eu. Then
f (g(x))=f (−(x−μ)2)=e−(x−μ)2 We could also have choseng(x)=(x−μ)2andf (u)=e−u.
Symmetry
The functionf (x)=x2satisfiesf (−x)=f (x), as indeed does any even powerx2n(withn an integer, positive or negative). So iff (−x)=f (x)for allxin the domain off, implying that the graph off issymmetric about they-axisas shown in Fig. 4, thenf is called an evenfunction.
On the other hand, any odd powerx2n+1(withnan integer) such asf (x)=x3satisfies f (−x)= −f (x). So iff (−x)= −f (x)for allx in the domain off, implying that the graph off issymmetric about the origin, as shown in Fig. 5, thenf is called anodd function.
Finally,f issymmetric abouta iff (a+x)=f (a−x)for allx. Thegraphoff is thensymmetric about the linex=aas in Fig. 6. In Sec. 4.6 we showed that the quadratic functionf (x)=ax2+bx+cis symmetric aboutx= −b/2a. The functiony =e−(x−μ)2 from Example 2(b) is symmetric aboutx=μ.
y
x f
⫺x x
y
x f
⫺x x
y
x f
a a⫹x a⫺x
Figure 4 Even function Figure 5 Odd function Figure 6 Symmetric aboutx=a P R O B L E M S F O R S E C T I O N 5 . 2
1. Show graphically how you find the graph ofy= 14x2+1/xby adding the graph of 1/xto the graph ofy= 14x2. Assumex >0.
2. Sketch the graphs of: (a)y=√
x−x (b)y=ex+e−x (c)y=e−x2+x
3. Iff (x) =3x−x3andg(x) = x3, compute:(f +g)(x),(f −g)(x),(f g)(x),(f/g)(x), f (g(1)), andg(f (1)).
4. Letf (x)=3x+7. Computef (f (x)). Find the valuex∗whenf (f (x∗))=100.
5. Compute ln(lne)and(lne)2. What do you notice? (This illustrates how, if we define the function f2byf2(x)=(f (x))2, then, in general,f2(x)=f (f (x)).)