# 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).

##### Definition

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.

Example

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.

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