So far we have discussed functions in algebra and its importance. The subject would be incomplete unless various types of functions are not talked about. So here we will discuss various types of functions. These functions are classified based on certain attributes such as, expressibility of the function ( if they can be explicit or implicit), mapping of elements between the domain and co-domain) and the way functions are graphically represented.

Let f : X $\rightarrow$ Y be a given function, where X is the domain and Y is the co-domain or X is the set of independent variable and Y is the set of dependent variable.
In a function if the dependent variable is fully expressed in terms of independent variable, then that function is known as explicit Function.
Like Y= f(X), where y = 3x2 + 4x -1 is the explicit function.

If f : X $\rightarrow$ Y is the given function. This is said to be ab implicit function if the dependent variable has not been expressed in terms of independent variable i.e.
F(x,y) = 0
or 3x2 + 3y2 = 6 is an implicit function.

We can easily convert an implicit function into an explicit function, but after converting it into an explicit function, it becomes complex and we get two different function branches that's why we can't do this generally.

If f : A $\rightarrow$ B is a function. Then f is said to be an one one function or injective function, if each element of the domain A is mapped with distinct(one and only one) element of co-domain B.

Prove that if f : R$\rightarrow$ R defined by f(x) = x + 1 is one one.
Given f : R$\rightarrow$ R, and f(x) = x + 1.
Let x1, x2 $\in$ R(domain), then
f(x1) = f(x2)
$\Rightarrow$ x1 + 1 = x2 + 1
$\Rightarrow$ x1 = x2.
Hence f is one one function.

Function f : A $\rightarrow$ B is said to be bijective if it is both one one and onto. It means for every element of B(co-domain) is mapped by only one element of A(domain).

The function from the set { 1,2,3, 4,5} to the set { 8,9,10,11,12} defined by f(x) = x + 7, is bijective function(i.e. one one and onto).
Let f : X $\rightarrow$ Y be a function.
If f(-x) = f(x), then above function is an even function.
If f(-x) = -f(x), then the function is an odd function.

Prove that
(a) f(x) = sinh x is odd function,
(b) f(x) = 4x2 + 1 is even function.
(a) f(x) = sinh x = $\frac{e^{x}- e^{-x}}{2}$

Replace x by -x, we get
f(-x) = sinh (-x) = $\frac{e^{-x}- e^{-(-x)}}{2}$

= $\frac{e^{-x}- e^{x}}{2}$

= - $\left [ \frac{e^{-x}- e^{x}}{2} \right ]$

= -f(x)
Function is odd function.

(b) f(x) = 4x2 + 1 , replece x by -x
or f(-x) = 4(-x)2 + 1
or f(-x) = 4x2 + 1 = f(x)
So f is even function.
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Let f : A $\rightarrow$ B be a function and if two or more than two elements of domain A is mapped to the same element of the set B then the function is said to be many to one.

A = $\left \{(1,4),(2,4),(3,5),(4,6) \right \}$ is many one function since
elements 1 and 2 mapped to the same element 4 under function f.
Let f : X $\rightarrow$ Y be a function. Then f is into if for every element of A there exists an element in B.
For into function, it is not compulsory for every element in B to has a pre-image in A.
Let f : X $\rightarrow$ Y be a function. If every element of co-domain(B) has its pre image in domain(A) under the function f, then the function is said to be onto function.

Example: Is f(x) = 4x -3 onto where f : R $\rightarrow$ R.

Given that f(x) = 4x -3 where domain is R and co-domain is also R. For
onto function, every element from co-domain has its pre image. So we
will try to solve this y = f(x) = 4x -3 relation for x.
So y = 4x -3
or x = $\frac{y + 3}{4 }$
so every element has its pre image in R(domain).

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