epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 63
EXAMPLE 2. The function/(x) = x^ has a unique inverse function.
Proof: We start by recaUing that two functions, / and g, are inverse of
each other if
fog{x)=f(g{x))
= x
g of{x) = gif{x)) = X
for all real numbers x (because/in this case is defined for all real numbers
and its range is the collection of all real numbers).
Part 1. The inverse function of/exists.
Because the function/is described by an algebraic expression, we will look
for an algebraic expression for its inverse, g.
The function g has to be such that:
fog{x)=f{g{x)) = x.
Therefore, using the definition of/we obtain
This implies that
{g{x)f = X.
We need to check that the function obtained in this way is really the
inverse function of/ Because
fog(x)=f(^) = {^'=x
g o/W = gi?^^) = ^
= ^
we can indeed conclude that g is the inverse function of/
Part 2. The inverse function of/is unique.
In this case, we can estabUsh the uniqueness of g in two ways:
a. The function g is unique because of the way it has been found and
defined.
b. Let us assume that there exists another function, h, that is the inverse
of/ Then, by definition of inverse,
for all real numbers x.
h o/(x) = X
f o h{x) = X