Page 265 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Inverses of Functions 251
Because we’re hoping to have g = f −1 , we know that for any x ∈ B =
R \{2}, g(x) must be the unique y ∈ A such that f (y) = x. Thus, to find
a formula for g(x), we solve for y in the equation f (y) = x. Filling in the
definition of f, we see that the equation we must solve is
1
+ 2 = x.
y
Solving this equation we get
1 1 1
+ 2 = x ⇒ = x − 2 ⇒ y = .
y y x − 2
Thus, we define g : B → A by the formula
1
g(x) = .
x − 2
Let’s check that g has the required properties. For any x ∈ A,wehave

1 1 1

g( f (x)) = g + 2 = = = x.
x 1/x + 2 − 2 1/x
Thus, g ◦ f = i A . Similarly, for any x ∈ B,

1 1
f (g(x)) = f = + 2 = x − 2 + 2 = x,
x − 2 1/(x − 2)
so f ◦ g = i B . Therefore, as we observed earlier, f must be one-to-one and
onto, and g = f −1 .
2. Imitating the solution to part 1, let’s try to find a function g : B → A such
that g ◦ f = i A and f ◦ g = i B . Because applying f to a number squares
the number and we want g to undo the effect of f, a reasonable guess would
√
be to let g(x) = x. Let’s see if this works.
For any x ∈ B we have
√ √ 2
f (g(x)) = f ( x) = ( x) = x,
so f ◦ g = i B . But for x ∈ A we have
√
2 2
g( f (x)) = g(x ) = x ,

2
and this is not always equal to x. For example, g( f (−3)) = (−3) =
√
9 = 3 =−3. Thus, g ◦ f = i A . This example illustrates that you must
check both f ◦ g = i B and g ◦ f = i A . It is possible for one to work but
not the other.