Page 57 - Calculus Demystified

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SOLUTION CHAPTER 1 Basics
Using the Basic Rule, we attempt to solve
(f ◦ f −1 )(t) = t.

Writing this out, we have
2
[f −1 (t)] = t.
But now there is a problem: we cannot solve this equation uniquely for f −1 (t).
√ √
We do not know whether f −1 (t) =+ t or f −1 (t) =− t. Thus f −1 is not
a well defined function. Therefore f is not invertible and f −1 does not exist.
Math Note: Thereisasimpledevicewhichoftenenablesustoobtainaninverse—
even in situations like Example 1.42. We change the domain of the function. This
idea is illustrated in the next example.

EXAMPLE 1.43
2
Define f :{s : s ≥ 0}→{t : t ≥ 0} by the formula f(s) = s . Find f −1 .

SOLUTION
We attempt to solve
−1
(f ◦ f )(t) = t.
Writing this out, we have
−1
f(f (t)) = t
or
2
[f −1 (t)] = t.
This looks like the same situation we had in Example 1.42. But in fact things
√
have improved. Now we know that f −1 (t) must be + t, because f −1 must
have range S ={s : s ≥ 0}. Thus f −1 :{t : t ≥ 0}→{s : s ≥ 0} is given by
√
f −1 (t) =+ t.
2
You Try It: The equation y = x +3x does not describe the graph of an invertible
function. Find a way to restrict the domain so that it is invertible.

Now we consider the graph of the inverse function. Suppose that f : S → T
is invertible and that (s, t) is a point on the graph of f . Then t = f(s) hence
s = f −1 (t) so that (t, s) is on the graph of f −1 . The geometrical connection
between the points (s, t) and (t, s) is exhibited in Fig. 1.47: they are reflections of
each other in the line y = x. We have discovered the following important principle:
The graph of f −1 is the reflection in the line y = x of the graph of f .

Refer to Fig. 1.48.