Invertible Functions

An invertible function is a function that can be reversed, so that the original input can be obtained back from the output. This topic is important because it connects the ideas of one-one and onto functions, inverse functions, composition of functions, and graphical interpretation.

A function \[f: X \to Y\] is defined to be invertible if there exists a function \[g: Y \to X\] such that:

Where \[I_X\] and \[I_Y\] are identity functions on sets X and Y. The function g is the inverse of f, denoted as \[f^{-1}\].

An inverse function takes
us back where we started 

A function is invertible if it is:

• One-one (injective): different inputs give different outputs.

• Onto (surjective): every element of the codomain has a pre-image in the domain.

• Bijective: both one-one and onto; this is the exact condition for invertibility.

• The inverse of a bijection is also a bijection.

• The inverse of an inverse function is the original function itself:

  \[(f^{-1})^{-1} = f\]

• If you apply a function and then its inverse, you get the original input:

  \[f(f^{-1}(x)) = x \quad \text{and} \quad f^{-1}(f(x)) = x\]

• Reversal Law of Inverses: If \[f: A \to B\] and \[g: B \to C\] are both bijections, then their composition \[g \circ f\] is also a bijection, and:

  \[(g \circ f)^{-1} = f^{-1} \circ g^{-1}\]

• Reflective Property: The graph of an inverse function \[f^{-1}\] is the exact reflection of the original function f across the diagonal line y = x.

To find the inverse of a function:

1. Write \[y = f(x)\].

2. Replace \[f(x)\] by \[y\] and solve for x in terms of y.

3. Interchange x and y.

4. The resulting expression gives \[f^{-1}(x)\], provided the function is invertible.

A Function is called a self-inverse function if its inverse is the exact same as the original function.

• Condition: \[(f \circ f)(x) = I_x = x\].

• Examples: \[f(x) = \frac{5}{x}\] and \[g(x) = 7 - x\].

Let f : N → Y be a function defined as f(x) = 4x + 3, where,
Y = { y ∈ N : y = 4x + 3 for some x ∈ N }. Show that f is invertible. Find the inverse.

Solution: Consider an arbitrary element y of Y. By the definition of Y, y = 4x + 3, for some x in the domain N. This shows that

\[x=\frac{(y-3)}{4}\]. Define g : Y → N by

\[g(y)=\frac{(y-3)}{4}\]. Now, gof(x) = g(f(x)) = g(4x + 3) = \[\frac{(4x+3-3)}{4}\] = x and

fog(y) = f(g(y)) \[=f\left({\frac{(y-3)}{4}}\right)={\frac{4\left(y-3\right)}{4}}+3\] = y − 3 + 3 = y.

This shows that gof = Iₙ and fog = Iᵧ, which implies that f is invertible and g is the inverse of f.

Let f: R → R be defined as f(x) = 10x + 7. Find the function g: R → R such that gof = fog = I_R.

Solution:
Given f: R → R is defined as f(x) = 10x + 7, for all x ∈ R.

Consider any arbitrary element y of R

As y ∈ R, \[\frac{y-7}{10}\] ∈ R

Let us define g : R → R by g(y) = \[\frac{y-7}{10}\], for all y ∈ R

As f: R → R and g: R → R, the composite functions gof and fog both exist.

These can be computed as:

(gof)(x) = g(f(x)) = g(10x + 7)
= \[\frac{(10x+7)-7}{10}\]= x, for all x ∈ R

and (fog)(y) = f(g(y)) = \[\frac{y-7}{10}\] = 10\[\frac{y-7}{10}\] + 7 = y, for all y ∈ R

⇒ gof = fog = I_R

Hence, the required function g: R → R is given by
g(y) = \[\frac{y-7}{10}\], for all y ∈ R.

If f: R → (−1, 1) is defined by f(x) = \[f(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\] is invertible, find f⁻¹.

Solution:

Let y = f(x), given f is invertible

⇒ \[\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{y}{1}\](Apply componendo and dividendo)

⇒ \[\frac{(e^x-e^{-x})+(e^x+e^{-x})}{(e^x-e^{-x})-(e^x+e^{-x})}=\frac{y+1}{y-1}\]

⇒ \[\frac{2e^x}{-2e^{-x}}=\frac{y+1}{y-1}\Rightarrow e^{2x}=\frac{1+y}{1-y}\]

⇒ \[2x=\log_{e}\frac{1+y}{1-y}\Rightarrow x=\frac{1}{2}\log\frac{1+y}{1-y}\]

⇒\[f^{-1}(y)=\frac{1}{2}\log\frac{1+y}{1-y}\]

⇒ f⁻¹(y) = 1/2 logₑ((1 + y)/(1 − y))

Hence,

\[f^{-1}(x)=\frac{1}{2}\log\frac{1+x}{1-x}\]

Graph the function and connect the points. Then graph the inverse. Identify the domain and range of each function.

Solution: Switching the x and y values in each ordered pair, the ordered pairs of the inverse functions are obtained as.

Plotting the points and connecting them, the graph of the given function and its inverse function are obtained as shown in Fig. 2.35. Clearly, the graph of the inverse of the given function is its reflection in the line y = x.

Their domain and range are as follows:

Given function:
Domain : {x : 0 ≤ x ≤ 8}
Range : {y : 2 ≤ y ≤ 7}

Inverse Function:
Domain : {x : 2 ≤ x ≤ 7}
Range : {y : 0 ≤ y ≤ 8}