Inverse Functions

Functions $f: X \to Y$ and $g: Y \to X$ are inverses if

$$f(g(y)) = y \quad\quad\hbox{and}\quad\quad g(f(x)) = x$$

for all $x \in X$ and $y \in Y$ . If f has an inverse, it is often denoted $f^{-1}$ . However, $f^{-1}$ does not mean "$\dfrac{1}{f}$ "!


Example. $f(x) = x^3$ and $g(x) = x^{1/3}$ are inverses, since

$$f(g(x)) = f(x^{1/3}) = (x^{1/3})^3 = x \quad\quad\hbox{and}\quad\quad g(f(x)) = g(x^3) = (x^3)^{1/3} = x$$

if x is a real number.

Notice that $f(x) = x^3$ but the inverse is not $\dfrac{1}{x^3}$ !


Example. Functions which are inverses "undo" one another. Thus, if f and $f^{-1}$ are inverses and f takes 4 to 17, then $f^{-1}$ must take 17 to 4.

In symbols,

$$f(4) = 17 \quad\hbox{implies}\quad f^{-1}(17) = 4.\quad\halmos$$


Example. In some cases, it's possible to find the inverse of a function algebraically. I'll find the inverse of $f(x) = x^3 + 5$ .

First, I'll write it as $y =
   x^3 + 5$ .

Next, switch x's and y's:

$$x = y^3 + 5.$$

(This means you should replace each "x" with a "y" and replace each "y" with an "x".)

Now solve for y in terms of x:

$$\eqalign{ x & = y^3 + 5 \cr x - 5 & = y^3 + 5 - 5 \cr x - 5 & = y^3 \cr (x - 5)^{1/3} & = (y^3)^{1/3} \cr (x - 5)^{1/3} & = y \cr}$$

Since I was able to solve for y in terms of x, the result is the inverse function: $f^{-1}(x)
   = y = (x - 5)^{1/3}$ .

The procedure I used tells something about the relation between the graphs of a function and its inverse. Since the inverse is obtained by swapping x's and y's, the graph of $f^{-1}$ is a mirror image of the graph of f across the line $y = x$ .

In the picture below, I've shown the graphs of $f(x) = x^3 + 5$ , $f^{-1}(x) = (x - 5)^{1/3}$ , and $y = x$ :

$$\hbox{\epsfysize=2in \epsffile{inverse-functions1.eps}}\quad\halmos$$


Example. Find the inverse of $f(x) = \dfrac{1}{x + 1}$ .

Let $y = \dfrac{1}{x + 1}$ . Swap x's and y's to obtain

$$x = \dfrac{1}{y + 1}.$$

Solve for y:

$$\eqalign{ x(y + 1) & = 1 \cr \dfrac{1}{x} \cdot x(y + 1) & = \dfrac{1}{x} \cdot 1 \cr y + 1 & = \dfrac{1}{x} \cr y + 1 - 1 & = \dfrac{1}{x} - 1 \cr y & = \dfrac{1}{x} - 1 \cr}$$

Therefore, $f^{-1}(x) = y =
   \dfrac{1}{x} - 1$ .


Example. Find the inverse of $f(x) = \dfrac{x - 5}{x}$ .

Let $y = \dfrac{x - 5}{x}$ . Swap x's and y's to obtain

$$x = \dfrac{y - 5}{y}.$$

Solve for y:

$$\eqalign{ x & = \dfrac{y - 5}{y} \cr y \cdot x & = y \cdot \dfrac{y - 5}{y} \cr xy & = y - 5 \cr xy - y & = y - 5 - y \cr xy - y & = -5 \cr y(x - 1) & = -5 \cr \dfrac{1}{x - 1} \cdot y(x - 1) & = \dfrac{1}{x - 1} \cdot -5 \cr y & = \dfrac{-5}{x - 1} \cr}$$

Hence, $f^{-1}(x) = y =
   \dfrac{-5}{x - 1}$ .


Not every function has an inverse. For example, consider $f(x) = x^2$ . Now $f(2) =
   4$ , so $f^{-1}$ should take 4 back to 2. But $f(-2) = 4$ as well, so apparently $f^{-1}$ should take 4 to -2. $f^{-1}$ can't do both, so there is no inverse! The problem is that you can't undo the effect of the squaring function in a unique way.

On the other hand, if I restrict $f(x) = x^2$ to $x \ge 0$ , then it has an inverse function: $f^{-1} = \sqrt{x}$ .

A function f is one-to-one or injective if different inputs go to different outputs:

$$x \ne y \quad\hbox{implies}\quad f(x) \ne f(y).$$


Example. $f(x) = x^4$ is not one-to-one, because different inputs can produce the same output. For example,

$$f(3) = 3^4 = 81 \quad\hbox{and}\quad f(-3) = (-3)^4 = 81.$$

On the other hand, $g(x) = 3x
   + 4$ is one-to-one. For suppose the inputs a and b produce the same output: $f(a) = f(b)$ . Then

$$3a + 4 = 3b + 4a.$$

Then

$$\matrix{& 3a & + & 4 & = & 3b & + & 4 \cr - & & & 4 & & & & 4 \cr \noalign{\vskip2pt\hrule\vskip2pt} & 3a & & & = & 3b & & \cr / & 3 & & & & 3 & & \cr \noalign{\vskip2pt\hrule\vskip2pt} & a & & & = & b & & \cr}$$

That is, the inputs a and b were the same to begin with.


A graph of a function represents a one-to-one function if every horizontal line hits the graph at most once.

$$\hbox{\epsfxsize=2.5in \epsffile{inverse-functions2.eps}} \hskip0.5in \hbox{\epsfxsize=2.5in \epsffile{inverse-functions3.eps}}$$

A one-to-one function has an inverse: Since a given output could have only come from one input, you can undo the effect of the function.


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