Here, you will learn what is inverse of a function, its properties and how to find the inverse of a function.
Letโs begin โ
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)ย The inverse of bijection is unique.
(b)ย The inverse of bijection is also a bijection.
(c)ย 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 andย 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.ย
Example : Let f : R \(\rightarrow\) R be defined by f(x) = \((e^x โ e^{-x})\)/2. Is f(x) invertible?. If so, find its inverse.
Solution : Let us check the invertibility of f(x) :
(a) One-One :
f(x) = \(1\over 2\)\((e^x โ e^{-x})\) \(\implies\) f'(x) = \(1\over 2\)\((e^x + e^{-x})\)
f'(x) > 0, f(x) is increasing function
\(\therefore\) ย 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 \(\implies\) \(e^{2y} โ 2xe^{y}\) โ 1 = 0
\(\implies\) ย \(e^y\) = \(2x \pm \sqrt{4x^2 + 4}\over 2\) \(\implies\) \(e^y\) = \(x \pm \sqrt{1 + x^2}\)
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 + \(\sqrt{1 + x^2}\))
