Question

the inverse function of the following real valued function of real variable
exist ? Give reasons.
f(x)= x​

Answer 1

SOLUTION :

GIVEN

A real valued function 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 is given by

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

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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LEARN MORE FROM BRAINLY

Let A={1,2,0,-1} B={1,3-1,-3} and

function f = A to B f(x) =px+q

If f={ (1,1),(2,3),(0,-1),(-1,-3)}

find the value of p & q

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