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between 0 and 150) Both functions have the same domain, and both take values in the non-negative integers But there is a fundamental difference between f and g If you are given a Social Security number, then you can determine the person to whom it belongs There will be one and only one person with that number But if you are given a number between 0 and 150, then there will probably be millions of people with that age You cannot identify a person by his/her age In summary, if you know g (x) then you generally cannot determine what x is But if you know f (x) then you can determine what x is This leads to the main idea of this subsection Let f : S T be a function We say that f has an inverse (is invertible) if there is a function f 1 : T S such that ( f f 1 )(t) = t for all t T and ( f 1 f )(s) = s for all s S Notice that the symbol f 1 denotes a new function which we call the inverse of f
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Basic Rule for Finding Inverses
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To find the inverse of a function f, we solve the equation ( f f 1 )(t) = t for the function f 1 (t)
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Find the inverse of the function f ( s) = 3s
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We solve the equation ( f f 1 ) ( t) = t This is the same as f ( f 1 ( t) ) = t We can rewrite the last line as 3 f 1 ( t) = t
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or f 1 ( t) = Thus f 1 ( t) = t/3 t 3
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EXAMPLE
Let f : R R be defined by f ( s) = 3s 5 Find f 1
SOLUTION
We solve ( f f 1 ) ( t) = t or f ( f 1 ( t) ) = t or 3[ f 1 ( t) ]5 = t or [ f 1 ( t) ]5 = or f 1 ( t) = t 3
YOU TRY IT Find the inverse of the function g( x) =
x 5
It is important to understand that some functions do not have inverses
EXAMPLE
Let f : R {t : t 0} be defined by f ( s) = s 2 If possible, find f 1
1 B A S I C S
SOLUTION
Using the Basic Rule, we attempt to solve ( f f 1 ) ( t) = t Writing this out, we have [ f 1 ( t) ]2 = 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 There is a simple device which often enables us to obtain an
inverse---even in situations like Example 142 We change the domain of the function This idea is illustrated in the next example EXAMPLE
Define f : {s : s 0} {t : t 0} by the formula f ( s) = s 2 Find f 1
SOLUTION
We attempt to solve ( f f 1 ) ( t) = t Writing this out, we have f ( f 1 ( t) ) = t or [ f 1 ( t) ]2 = t This looks like the same situation we had in Example 142 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
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YOU TRY IT The equation y = x2 + 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 Figure 147: 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 reflectionin the line y = x of the graph of f Refer to Figure 148
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