# 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 + $$\sqrt{1 + x^2}$$)