MA101.4 Inverse Functions

Let f be a function and let y=f(x), x \in D(f). We often call x the independent variable and y the dependent variable. Now consider specifying a number y \in R(f) and working out what x’s are in the domain of the f, for which f(x)=y. There may of course be many such x’s corresponding to a particular y, but functions for which there is only ever one such x have a special name, they are called injections (their old name was one-to-one).


A function is an injection if for all x,y f(x)=f(y) \Rightarrow x=y

Or if you prefer \forall x,y:\; x \neq y \Rightarrow f(x) \neq f(y)

(\forall – for all and \Rightarrow – implies.)

Injections are easily recognised from their graphs in that any line parallel to the x axis will meet the graph in at most one point.

Let f be an injection, then we use f^{-1}(y), y \in R(y) to denote that single element x \in D(f) which is such that f(x)=y.

The function f^{-1} : R(f) \rightarrow D(f) is called the inverse function of the f.

By the definition of f^{-1} we have, D(f^{-1}) = R(f).

It is also clear that R(f^{-1}) = D(f).

Note that functions that are not injections do not have inverse functions.


Find the inverse function of the function f given by:

f(x) = \begin{cases} (1 - x)^2 &\mbox{}0 \leq x \leq 1 \\ x+1 &\mbox{} 1 < x < 2 \end{cases}

Solution: To have a sketch is a great help

D(f)=[0,2) and R(f)=[0,1] \cup (2,3)

So D(f^{-1}) = R(f) = [0,1] \cup (2,3).

Since the rule for f is given in two pieces we find the rule for f^{-1} in two pieces.

Let\; y \in [0,1] \\ So,\\ y=(1-x)^2 \\ y = 1 - 2x +x^2\\ x^2 - 2x + 1 - y = 0 \\ x = \frac{2-\sqrt{4-4(1-y)}}{2} \\ x = 1 - \sqrt{y}

Let\; y \in (2,3) \\So,\\ y = x+1 \; thus \; x = y-1.

Whence f^{-1} is given by

f^{-1}(y) = \begin{cases} 1-\sqrt{y} &\mbox{}0 \leq y \leq 1 \\ y-1 &\mbox{} 2 < y < 3 \end{cases}

Note that f is an injection and so f^{-1} exists then the graphs f and f^{-1} will be reflections of one another in the line y = x.