What is Inverse of a Function – Properties and Example

Here, you will learn what is inverse of a function, its properties and how to find the inverse of a function.

Let’s begin –

Inverse of a Function

Let f : A \(\rightarrow\) B be a one-one & onto function, then there exists a unique function g : B \(\rightarrow\) A such that f(x) = y \(\iff\) g(y) = x, \(\forall\) x \(\in\) A & y \(\in\) B. Then g is said to be inverse of f.

Properties of inverse functions

(a)  The inverse of bijection is unique.

(b)  The inverse of bijection is also a bijection.

(c)  If f & g are two bijections f : A \(\rightarrow\) B, g : B \(\rightarrow\) C then the inverse of gof exists and \((gof)^{-1}\) = \(f^{-1}\)o\(g^{-1}\).

How to Find Inverse of Function

In order to find the inverse of a function, we may use the following algorithm.

Let f : A \(\rightarrow\) B be a bijection. To find the inverse of f we follow the following steps:

Step 1 : Put f(x) = y, where y \(\in\) B and  x \(\in\) A.

Step 2 : Solve f(x) = y to obtain x in terms of y.

Step 3 : In the relation obtained in step 2 replace x by \(f^{-1}(y)\) to obtain the required inverse of f. 

Example : Let f : R \(\rightarrow\) R be defined by f(x) = \((e^x – e^{-x})\)/2. Is f(x) invertible?. If so, find its inverse.

Solution : Let us check the invertibility of f(x) :

(a) One-One :

f(x) = \(1\over 2\)\((e^x – e^{-x})\) \(\implies\) f'(x) = \(1\over 2\)\((e^x + e^{-x})\)

f'(x) > 0, f(x) is increasing function
\(\therefore\)   f(x) is one-one function.

(b) Onto :

As x tends to larger and larger values so does f(x) and when x \(\rightarrow\) \(\infty\), f(x) \(\rightarrow\) \(\infty\)
Similarly as x \(\rightarrow\) -\(\infty\), f(x) \(\rightarrow\) -\(\infty\) i.e. -\(\infty\) < f(x) < \(\infty\) so long as x \(\in\) (-\(\infty\), \(\infty\))
Hence the range of f is same as the set R. Therefore f(x) is onto.

Since f(x) is both one-one and onto, f(x) is invertible.

(c) To find \(f^{-1}\)(x) : Interchange x & y

\(1\over 2\)\((e^y + e^{-y})\) = x \(\implies\) \(e^{2y} – 2xe^{y}\) – 1 = 0

\(\implies\)   \(e^y\) = \(2x \pm \sqrt{4x^2 + 4}\over 2\) \(\implies\) \(e^y\) = \(x \pm \sqrt{1 + x^2}\)

Since \(e^y\) > 0, hence negative sign is ruled out and Hence \(e^y\) = \(x + \sqrt{1 + x^2}\)

Taking logarithm, we have y = ln(x + \(\sqrt{1 + x^2}\)) or \(f^{-1}\)(x) = ln(x + \(\sqrt{1 + x^2}\))

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