1.4 Inverses of Functions

Roughly speaking, an inverse for the function f is a function that undoes the action of f. Applying f first and then the inverse restores every mem­ber of the domain of f to where it started. See Figure 1.

example 1 (a) The functions x² and with domains [0, ) are “inverses” to

each other. If you apply these operations in either order to some value,

the original value is obtained. Try it on a calculator! In symbols,

= x and ( )² = x for x  [0, ).

(b) The function 1/x is its own “inverse.” If you apply the operation

twice to some value, you get the original value. That is,

= x for all nonzero x in R. ■

Here is the precise definition. An inverse of a function f: S  T is a function f^--1: T  S such that f ^-1 o f = l_S and f o f ^-1 = l_T, i.e., such that

f ^-1(f(x)) = x for all x  S

and

f( f ^–1(y)) = y for all y  T.

Not all functions have inverses; those that do are called invertible func­tions. We will see in the proof of the theorem that the defining conditions completely determine f ^-1 if it exists, so an invertible function can't have two different inverses.

example 2 Consider a positive real number b with b  1. The important examples

of b are 2, 10 and the number e that appears in calculus and is

approxi­mately 2.718. The function f_b given by f_b(x) == b^x for x  R

has an inverse with domain (0, ), which is called a logarithm

function. We write (y) = log_b y, by the definition of an inverse

we have

log_b b^x = x for every x  R

and

b^{logb y}= y for every y  (0, ).

In particular, e^x and log_e x are inverse functions. The function log_e x is called the natural logarithm and is often denoted In x. The functions 10^x and log₁₀ x are inverses, and so are 2^x and log₂ x. The functions log₁₀ x = log and log_e x = ln appear on many calculators; such calculators also allow one to compute the inverses 10^x and e^x of these functions. To com­pute log₂ x on a calculator, use one of the formulas

or

■

The next theorem tells us which functions are invertible.

The function f: S  T is invertible if and only if f is one-to-one and

maps S onto T.

Theorem

Proof. Suppose that f has an inverse, f ^-1. If x₁, x₂  S with f(x₁)=f(x₂), then

x₁=f ^–1(f(x₁))=f ^–1(f(x₂))=x₂.

Thus f is one-to-one. Moreover, if y  T, then f ^-1(y) belongs to S and f(f ^–1(y))=y; so y  Im(f). Hence T = Im(f) and f maps S onto T.

Conversely, if f maps S onto T, then for each y in T there is some x  S with f(x) = y. If f is also one-to-one, then there is exactly one such x, and we get a formula for f ^-1, namely:

(*) f ^–1( y) = that unique x  S such that f(x) = y.

This definition immediately gives f(f ^-1 (y)) = y, and f^-1(f(x)) is the unique member of S that f maps to f(x), namely x itself. Thus f ^-1, as defined by (*), meets the conditions to be the inverse of f. ■

This proof also shows how to get f ^-1(y) if f is invertible. Simply solve for x in terms of y.

EXAMPLE 3 Consider the function f: R  R given by f(x) = x³ + 1. To see that f

is one-to-one, we note that

f(x₁) = f(x₂) implies x₁³ + 1 == x₂³ + 1

implies x₁³ = x₂³ implies x₁ = x₂;

the last implication holds because each real number has a unique cube root.

To check that f maps R onto R, consider y in R. We need to find x  R so that f(x) = y, i.e., we need to solve x³ + 1 == y for x. When we do, we get x = , which belongs to R. Hence f maps R onto R.

Since f is one-to-one and maps R onto R, f is invertible, by the theorem. We found f ^-1 when we solved for x. Thus f ^-1(y) = . This formula makes sense for each y in R, and so f ^-1 is completely determined. ■

EXAMPLE 4 Consider the function g: ZZ  ZZ given by g(m, n) = (-n,-m).

We will check that g is one-to-one and onto, and then we'll find its inverse. To show that g is one-to-one we need to show that

g(m, n) = g(m', n') implies (m, n) == (m', n').

If g(m, n) = g(m',n) then (-n, -m) =(-n,-m). Since these ordered pairs are equal we must have –n =  n' and –m = m'. Hence m = m' and n = n' so that (m, n) = (m', n'), as desired.

To show that g maps onto Z  Z, consider (p, q) in Z  Z. We need to find (m, n) in Z  Z so that g{m, n) == (p, q). Thus we need (n, m) == (p, 4), and this tells us that n should be p and m should be q. In other words, given (p, q) in Z  Z we see that (q,p) is an element in Z  Z such that g(q, p)= (p, q). Thus g maps Z  Z onto Z  Z.

To find the inverse of g we need to take (p, q) in Z  Z and find g^-1(p, q). We just did this in the last paragraph; g maps (q, p) onto (p, q), and hence g^-1(p, q) =(-q, -p) for all (p, q) in Z  Z.

It is interesting to note that g = g ^-1 in this case. ■

Inverses of functions are so useful that we sometimes restrict func­tions that are not one-to-one to smaller domains on which they are one-to-one. If we then arrange for the codomain to equal the image of the function, we obtain an invertible function.

EXAMPLE 5 (a) Consider : R  R defined by (x) = x². Then  is not one-to-

one, but it is one-to-one if we restrict the domain to [0, ). Thus we define a new function F by the same rule, F(x) = x², but with Dom(F) == [0, ). Then F is one-to-one. In fact, F: [0, )  [0, ) is one-to-one and onto. It is this function that has F^-1(x) = as its inverse; see Example l(a).

The function F is called the restriction of  to [0, ). This sort of restriction is clearly possible and desirable in many settings of interest.

(b) You should be able to follow this example even if you know no trigonometry. It turns out that none of the trigonometric functions are one-to-one. For example, consider the graph of sin x in Figure 2. Never­theless, sin x is one-to-one if its domain is restricted to, say, [  /2, /2].

FIGURE 2

See Figure 3(a), where we have denoted the restriction by Sin x. With codomain [-1,1], we obtain an invertible function; the inverse is given in Figure 3(b). This is the inverse sine or arc sin encountered in trigonom­etry, calculus and many hand-held calculators. ■