Question

Does the inverse function of the following real valued function of real variable
cxist ? Give reasons.
f(x)=1*​

Answer 1

Given : f(x)=[x]

To Find : Does the inverse function  of real variable exist

Solution:

f(x)=[x]

[x]   = Greatest integer function

x = 1  => f(1)  = 1

x= 1.01  => f(1.01) = 1

x = 1.2 => f(1.2) = 1

x = 1.99 => f(1.99) = 1

f(1) =  f(1.2) but 1#1.2  Hence function is not one to one

As function is not one to one  so function  is not bijective  

A function is said to be invertible iff it is bijective

Since  function   is not bijective  

Hence the inverse does not exists

=>  the inverse function of the f(x)=[x] of real variable does not exist

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क्या निम्नलिखित वास्तविक चर के वास्तविक-मानित फलन का प्रतिलोम फलन प्राप्त होग

कारण  f(x)=|x|

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Answer 2

SOLUTION :

COMPLETE QUESTION

Does the inverse function of the following real valued function of real variable exist ? Give reasons.

f(x)= x

GIVEN

A real valued function f of Real variables is defined as

$\sf{ f(x) = x\: }$

TO DETERMINE

The reason of existence of inverse of f(x)

CONCEPT TO BE IMPLEMENTED

For a function f(x) the inverse of f(x) exists if f(x) is bijective

EVALUATION

Here

A real valued function f of Real variables is defined as

$\sf{ f(x) = x\: }$

CHECKING FOR INJECTIVE

$\sf{Let \: x_1, x_2 \in \mathbb{R} \: such \: that \: \: f (x_1)=f(x_2) \: }$

Now

$\sf{ f (x_1)=f(x_2) \: \: gives }$

$\sf{x_1 = x_2 \: \: \: \: \: (\because \: f(x) = x \: })$

So f(x) is injective

CHECKING FOR SURJECTIVE

$\sf{Let \: \: y \in \mathbb{R} \: \: \: (co - domain \: set \: )}$

If possible let there exists a real number such that

$\sf{ y = f(x)\: }$

$\implies \sf{ y = x\: }$

So $\sf{ f(y) = y\: }$

Now y is arbitrary

So For every element y in the co-domain set there exists an element in domain set such that

$\sf{f(y) = y \: }$

So f(x) is surjective

Hence f(x) is bijective

Therefore

$\sf{ {f}^{ - 1} \: \: \: exists}$

━━━━━━━━━━━━

ADDITIONAL INFORMATION

$\sf{ let \: \: {f}^{ - 1}(x) = y \: \: \:}$

$\implies \sf{f(y) = x \: }$

$\implies \sf{y = x \: }$

Hence $\sf{ {f}^{ - 1} (x) = x \: }$

━━━━━━━━━━━━━━━━

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