If f is a function defined as y = f(x), then inverse function of f is x = f ^{-1}(y) i.e. f^{-1 }defined from y to x. In inverse function co-domain of f is the domain of f ^{-1} and the domain of f is the co-domain of f^{ -1}.

Only one-to-one functions has its inverse since these functions has one to one correspondences i.e. each elements from the range correspond to one and only one domain element.

Let a function f : A$\rightarrow$ B is defined, then f is said to be invertible if there exists a function g : B $\rightarrow$ A in such a way that if we operate f{g(x)} or g{f(x)} we get the starting point or value.

**Example****:** Show that the function f : X $\rightarrow $ Y, such that f(x)= 5x + 7,

$\forall$ x,y $\in$ N is an invertible.

**Solution****:** Let y$\in$ N $\rightarrow$ y = f(x) = 5x + 7 for x$\in$N

$\Rightarrow$ x = $\frac {y-7} { 5} $.

If we define h : Y $\rightarrow$ X by h(y) = $\frac {y-7} { 5} $.

Again hof(x) = h[ f(x) ] = h{ 5x + 7 } = $\frac {5(y-7)} { 5} $ + 7 = x

And foh(y) = f[ h(y) ] = f( $\frac {y-7} { 5} $) = $\frac {5(y-7)} { 5} $ + 7 = y

Hence f is invertible function and h is the inverse of f.