A function is defined as a relation between the set of inputs called domain and the set of outputs called range. It is a special type of relation between the input value and the output value. It is in which each element of the domain is paired with one or more element of the range. Mapping of function in mathematics is defined as a mathematical relation such that one or more elements of a given set called domain is associated with one or more elements of another set which is the range of the function. Suppose there are two sets called A and B each containing certain elements, then the different types of function mapping is as follows: • One to one function – If each element of set A is related or mapped with different elements of set B, then that type of function mapping is called one to one function or injective function. • Many to one function – If there are two or more elements of set A related or mapped with a particular element of set B, then that type of function mapping is called many to one function. • Onto function – When all the elements of set B is connected to the elements of set A such that there are no such element of set B which is not connected or mapped then that type of function mapping is called onto function or surjective function. • Into function – If there are few idle elements in set B one or two elements in set B which is not connected to any elements of set A, then that type of function mapping is called into function. • One to one onto function – A function having properties of both one to one function and onto function, then that type of function mapping is called one to one onto function.

If a function has a domain $A$ and codomain $B$, then its inverse function would have the domain $B$ and codomain $A$. A function and its inverse function have its domain and range interchanged. The symbol to denote an inverse function is '$f  1$'. Basic properties of inverse function shall include the following:
• For a function having its domain as $A$ and range as $B$, would have its inverse satisfy
$F^{1}\ [f(x)]$ = $x$ for every $x$ in $A$ and $f[f^{1}\ (x)]$ = $x$ for every $x$ in $B$
• A function $f$ has its inverse as $f^{1}$. Then, $f$ and $f^{1}$ are inverse of each other
• A function and its inverse shall have its domain and range interchanged, so we can say that
Domain of $f$ = Range of its inverse $f^{1}$
Domain of inverse of function $f^{1}$ = Range of function of $f$
• The graph of the inverse function $f^{1}$ can be drawn given the graph of the function $f$ by reflecting it over the line $y$ = $x$
• The restriction of an inverse function is that it has to be an one to one function
Steps to find out the inverse of a function, given the one to one function are as follows:
• Step 1  We need to replace $f(x)$ with $y$
• Step 2  We need to interchange $x$ and $y$
• Step 3  We then solve the equation for $y$
The $y$ solved is actually the resulting equation in $f^{1}(x)$
Tests can be conducted to check whether a given function is a one to one function or not. A function must pass both the vertical line test and horizontal line test to call it as one to one function.
On drawing a function on a graph if a horizontal line is drawn across it and it crosses or cuts the graph of the given function at a single point, then it is called a one to one function. If the horizontal line cuts the given function at more than one point, then we cannot call it as an injective function. It means for a single $x$ value of the coordinate, there should be a single value of $y$ to be a one to one function. This is called the horizontal line test.
A one to one function need to be check whether it is a function or not, and for that we carry out the vertical line test. So, vertical line test is conducted to see whether a given injective function is actually a function or not. On drawing a function on a graph if a vertical line is drawn across it and it crosses or cuts the graph of the given function at a single point, then it is called a function. If the vertical line cuts the given function at more than one point, then we cannot call it as function. It means for a single $y$ value of the coordinate, there should be a single value of $x$ to be a function. This is called the vertical line test.
When two or more elements of set $A$ have the same image of an element in set $B$, then that type of mapping of function is called many to one function. A many to one function can be checked by drawing it on a graph and then a horizontal line is drawn across it. If the horizontal line cuts the graph of a function at more than a single point it means for different values of $x$ coordinate the $y$ coordinate remains the same. In many to one function there are various values of $x$ coordinate but the $y$ coordinate remains constant.
Example 1:
Is function $f$ defined by $[(1, 2),\ (3, 4),\ (5, 6),\ (7, 8)]$ a one to one function?
Solution:
The $x$ coordinates given are $1, 3, 5$ and $7$. For each value of the domain there a single unique image in the range. The values of the range given are $2, 4, 6$ and $8$. So, for every $x$ value there is a corresponding value of $y$. Therefore, the given function $f$ is defined as a one to one function.
Example 2:
Is function $f$ defined by $[(2, 2),\ (3, 4),\ (5, 4),\ (1, 4)]$ a many to one function?
Solution:
The $x$ coordinates given are $1, 2, 3$ and $5$. We see that there is repetition of $y$ values in the range. There is three times number $4$ given in the range. Number $3$ is paired with $4$, number $5$ is paired with $4$ and number $1$ is paired with $4$. So, we can call the given function as many to one function.
Example 3:
Determine whether the given function is one to one function or not:
$F(x)$ = $8x  3$
Solution:
Function f is said to be a one to one function if $f(x1)$ = $f(x2)$ implies that $x1$ = $x2$.
$F(x1)$ = $f(x2)$
$8x1  3$ = $8x2  3$
Adding $3$ on both sides eliminates the $3$ from both sides, giving us
$8x1  3 + 3$ = $8x2  3 + 3$
$8x1$ = $8x2$
Dividing both sides by $8$, we get
$\frac{8x1}{8}$ = $\frac{8x2}{8}$
$X1$ = $x2$
Thus, $f$ is a one to one function.
Example 4:
Find the inverse of $f$:
a) $f(x)$ = $2x  8$
b) $f(x)$ = $\frac{x + 4}{5x  9}$
Solution:
a) Finding the inverse of $f(x)$ = $2x  8$
Step 1: We need to replace $f(x)$ with $y$
$Y$ = $2x  8$
Step 2: We need to interchange $x$ and $y$
$X$ = $2y  8$
Step 3: We then solve the equation for $y$
$X$ = $2y  8$
Adding $8$ on both sides
$X + 8$ = $2y$
Dividing both sides by $2$ to isolate $y$
$\frac{X + 8}{2}$ = $\frac{2y}{2}$
$\frac{X + 8}{2}$ = $y$
The $y$ solved is actually the resulting equation in $f^{1}(x)$
$f^{1}(x)$ = $\frac{x + 8}{2}$
b) Finding the inverse of $f(x)$ = $\frac{x + 4}{5x  9}$
Step 1: We need to replace $f(x)$ with $y$
$f(x)$ = $\frac{x + 4}{5x  9}$
$y$ = $\frac{x + 4}{5x  9}$
Step 2: We need to interchange $x$ and $y$
$y$ = $\frac{x + 4}{5x  9}$
$x$ = $\frac{y + 4}{5y  9}$
Step 3: We then solve the equation for $y$
$x$ = $\frac{y + 4}{5y  9}$
On cross multiplication
$x \times (5y  9)$ = $y + 4$
$5xy  9x$ = $y + 4$
Adding $9x$ on both sides
$5xy  9x + 9x$ = $y + 4 + 9x$
$5xy$ = $y + 4 + 9x$
Subtracting $y$ on both sides
$5xy  y$ = $y + 4 + 9x  y$
$5xy  y$ = $4 + 9x$
Factor out $y$
$y(5x  1)$ = $4 + 9x$
Isolating $y$ by dividing both sides by $5x  1$
$y$ = $\frac{(4 + 9x)}{(5x  1)}$
The $y$ solved is actually the resulting equation in $f^{1}(x)$
$f^{1}(x)$ = $\frac{(4 + 9x)}{(5x  1)}$