1. Introducción
1.2 Antecedentes del tema
Suppose the demand for a commodity is a function x of its price p. Suppose that price p is not constant, but depends on time t. Then it is natural to regard x as a function of t.
In general, if y is a function of u, and u is a function of x, then y can be regarded as a function of x. We call y a composite function of x. If we denote the two functions involved by f and g, with y= f (u) and u = g(x), then we can replace u by g(x) and so write y in the form
y= f!g(x)"
Note that when computing f!g(x)", we first apply g to x to obtain g(x), and then we apply f to g(x). The operation of first applying g to x and then f to g(x) defines a composite
function. Here g(x) is called the kernel, or interior function, while f is called the exterior function.
NOTE 1 The function that maps x to f!g(x)"is often denoted by f ' g. This is read as
“f of g” or “f after g”, and is called the composition of f with g. Correspondingly, g' f denotes the function that maps x to g!f (x)". Thus, we have
(f' g)(x) = f!g(x)" and (g' f )(x) = g!f (x)"
Usually, f ' g and g ' f are quite different functions. For instance, if g(x) = 2 − x2 and
f (u)= u3, then (f ' g)(x) = (2 − x2)3, whereas (g' f )(x) = 2 − (x3)2 = 2 − x6; the two resulting polynomials are not the same.
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It is easy to confuse f' g with f · g, especially typographically. But these two functions are defined in entirely different ways. When we evaluate f ' g at x, we first compute g(x)and then evaluate f at g(x). On the other hand, the product f · g of f and g is the function whose value at a particular number x is simply the product of f (x) and g(x), so (f· g)(x) = f (x) · g(x).
Many calculators have several built-in functions. When we enter a number x0 and
press the key for the function f , we obtain f (x0). When we compute a composite
function given f and g, and try to obtain the value of f!g(x)", we proceed in a similar manner: enter the number x0, then press the g key to get g(x0), and again press the f key
to get f (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)=√xand g(x)= 1/x, so f (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 compute x3 + x2, which gives the interior
function, g(x)= x3 + x2. Then take the fiftieth power of the result, so the exterior
function is f (u)= u50. Hence,
f (g(x))= f (x3+ x2)= (x3+ x2)50
(b) We can choose the interior function as g(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 chosen g(x)= (x − µ)2and f (u)= e−u.
Symmetry
The function f (x)= x2satisfies f (−x) = f (x), as indeed does any even power x2n(with n
an integer, positive or negative). So if f (−x) = f (x) for all x in the domain of f , implying that the graph of f is symmetric about the y-axis as shown in Fig. 4, then f is called an
On the other hand, any odd power x2n+1(with n an integer) such as f (x)= x3satisfies
f (−x) = −f (x). So if f (−x) = −f (x) for all x in the domain of f , implying that the graph of f is symmetric about the origin, as shown in Fig. 5, then f is called an odd function.
Finally, f is symmetric about a if f (a+ x) = f (a − x) for all x. The graph of f is then symmetric about the line x= a as in Fig. 6. In Sec. 4.6 we showed that the quadratic function f (x)= ax2+ bx + c is symmetric about x = −b/2a. The function y = e−(x−µ)2
from Example 2(b) is symmetric about x= 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 about x= 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 of y= 1
4x2+ 1/x by adding the graph of 1/x to the graph of y= 1
4x2. Assume x > 0.
2. Sketch the graphs of: (a) y=√x− x (b) y = ex+ e−x (c) y= e−x2+ x
3. If f (x) = 3x − x3and g(x)= x3, compute: (f + g)(x), (f − g)(x), (fg)(x), (f/g)(x), f (g(1)), and g(f (1)).
4. Let f (x)= 3x + 7. Compute f (f (x)). Find the value x∗when f (f (x∗))= 100.
5. Compute ln(ln e) and (ln e)2. What do you notice? (This illustrates how, if we define the function f2by f2(x)= (f (x))2, then, in general, f2(x)̸= f (f (x)).)
5.3 Inverse Functions
Suppose that the demand quantity D for a commodity depends on the price per unit P according to
D= 30 P1/3
This formula tells us directly the demand D corresponding to a given price P . If, for example, P = 27, then D = 30/271/3 = 10. So D is a function of P . That is, D = f (P ) with f (P )= 30/P1/3. Note that demand decreases as the price increases.
If we look at the matter from a producer’s point of view, however, it may be more natural to treat output as something it can choose and consider the resulting price. The producer wants to know the inverse function, in which price depends on the quantity sold.
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This functional relationship is obtained by solving D = 30/P1/3 for P . First we obtain
P1/3= 30/D and then (P1/3)3= (30/D)3, so that the original equation is equivalent to P = 27 000
D3
This equation gives us directly the price P corresponding to a given output D. For example, if D = 10, then P = 27 000/103 = 27. In this case, P is a function g(D) of D, with
g(D)= 27 000/D3.
The two variables D and P in this example are related in a way that allows each to be regarded as a function of the other. In fact, the two functions
f (P )= 30p−1/3 and g(D)= 27 000D−3
are inverses of each other. We say that f is the inverse of g, and that g is the inverse of f . Note that the two functions f and g convey exactly the same information. For example, the fact that demand is 10 at price 27 can be expressed using either f or g:
f (27)= 10 or g(10)= 27
In Example 4.5.3 we considered an even simpler demand function D= 100 − P . Solving for P we get P = 100 − D, which was referred to as the inverse demand function.
Suppose in general that f is a function with domain Df = A, meaning that to each x in
Athere corresponds a unique number f (x). Recall that if f has domain A, then the range of f is the set B = Rf = {f (x) : x ∈ A}, which is also denoted by f (A). The range B
consists of all numbers f (x) obtained by letting x vary in A. Furthermore, f is said to be
one-to-one in A if f never has the same value at any two different points in A. In other
words, for each one y in B, there is exactly one x in A such that y= f (x). Equivalently, f is one-to-one in A provided that it has the property that, whenever x1and x2both lie in A
and x1 ̸= x2, then f (x1)̸= f (x2). It is evident that if a function is strictly increasing in all
of A, or strictly decreasing in all of A, then it is one-to-one. A particular one-to-one function f is illustrated in Fig. 1; another function g that is not one-to-one is shown in Fig. 2.
f f(A) ! B A y x g x1 y1 x2 A y x
Figure 1 f is one-to-one with domain A and
range B. f has an inverse
Figure 2 gis not one-to-one and hence
has no inverse over A. Which x-value should be associated with y1?
D E F I N I T I O N O F I N V E R S E F U N C T I O N
Let f be a function with domain A and range B. If and only if f is one-to-one, it has an inverse function g with domain B and range A. The function g is given by the following rule: For each y in B, the value g(y) is the unique number x in Asuch that f (x)= y. Then
g(y)= x ⇐⇒ y = f (x) (x∈ A, y ∈ B)
(1)
A direct implication of (1) is that
g(f (x))= x for all x in A and f (g(y))= y for all y in B (2) The equation g(f (x))= x shows what happens if we first apply f to x and then apply g to f (x): we get x back because g undoes what f did to x. Note that if g is the inverse of a function f , then f is also the inverse of g.
NOTE 1 If g is the inverse of f , it is standard to use the notation f−1for g. This some-
times leads to confusion. If a is a number, then a−1 means 1/a. But f−1(x)does not
mean 1/f (x)= (f (x))−1. For example: The functions defined by y= 1/(x2+ 2x + 3) and
y = x2+ 2x + 3 are not inverses of each other, but reciprocals.
In simple cases, we can use the same method as in the introductory example to find the inverse of a given function (and hence automatically verify that the inverse exists). Some more examples follow.
E X A M P L E 1 Solve the following equations for x and find the corresponding inverse functions: (a) y= 4x − 3 (b) y= √5 x+ 1 (c) y= 3x− 1
x+ 4
Solution:
(a) Solving the equation for x, we have the following equivalences for all x and all y: y= 4x − 3 ⇐⇒ 4x = y + 3 ⇐⇒ x = 14y+34
We conclude that f (x)= 4x − 3 and g(y) = 14y+34 are inverses of each other.
(b) We begin by raising each side to the fifth power and so obtain the equivalences y= √5 x+ 1 ⇐⇒ y5 = x + 1 ⇐⇒ x = y5− 1
These are valid for all x and all y. Hence, we have shown that f (x)= √5 x+ 1 and
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(c) Here we begin by multiplying both sides of the equation by x+ 4 to obtain y(x + 4) = 3x− 1. From this equation, we obtain yx + 4y = 3x − 1 or x(3 − y) = 4y + 1. Hence,
x = 4y+ 1 3− y
We conclude that f (x)= (3x − 1)/(x + 4) and g(y) = (4y + 1)/(3 − y) are inverses of each other. Observe that f is only defined for x ̸= −4, and g is only defined for y̸= 3. So the equivalence in (1) is valid only with these restrictions.