text: # Invertible Function
  
  Invertible Function
  
  An invertible function is a function that can be reversed or inverted by another function. In other words, if f(x) is an invertible function, there exists another function g(x) such that g(f(x)) = x for all x in the domain of f(x) and f(g(x)) = x for all x in the domain of g(x).
  
  The graph of an invertible function has the property that it is symmetric with respect to the line y = x. That is, if you reflect the graph of f(x) across the line y = x, you get the graph of its inverse function g(x).
  
  A function is invertible if and only if it is one-to-one (injective) and onto (surjective). In other words, every element in the range of f(x) corresponds to exactly one element in the domain of f(x), and every element in the domain of f(x) has a corresponding element in the range of f(x).
  
  For example,
  
  the function f(x) = 2x + 3 is invertible because it is both one-to-one and onto. Its inverse function is g(x) = (x - 3)/2.
  
  Properties of Invertible Function
  
  (i) A function f : X → Y is defined to be invertible, if there exists a function g : Y → X such that g o f = Ix and f o g = IY. The function g is called the inverse of f and is denoted by f –1.
  
  (ii) A function f : X → Y is invertible if and only if f is a bijective function.
  
  (iii) If f : X → Y, g : Y → Z and h : Z → S are functions, then h o (g o f) = (h o g) o f.
  
  (iv) Let f : X → Y and g : Y → Z be two invertible functions. Then g o f is also invertible with (g o f)–1 = f –1 o g–1.
  
  Example
  
  Let R be the set of real numbers and f : R → R be the function defined by f (x) = 3x + 5. Show that f is invertible and find f –1.
  Solution
  
  Here the function f : R → R is defined as f (x) = 3x + 5 = y (say).
  
  Then         3x = y – 5 or x =  $\frac{y-5}{3}$y53
  
  This leads to a function g : R → R defined as
  g (y) =  $\frac{y-5}{3}$y53
  Therefore, ( g o f ) (x) = g(f (x) = g (3x + 5)
  =$\frac{\text{3x+5-5}}{3}$3x+5-53   =x
  or        g o f = IR
  Similarly (f o g) (y) = f (g(y))
  = f( $\frac{y-5}{3}$y53  )  =  $3\left(\frac{y-5}{3}\right)+5$3(y53 )+5  = y
  or f o g = IR .
  Hence f is invertible and f –1 = g which is given by
  f –1 (x) =  $\frac{x-5}{3}$x53
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