Which statement could be used to explain why f(x) = 2x – 3 has an inverse relation that is a function?
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Any function f(x) will be an inverse function only if it is an injective as well as surjective function.
I mean, conditions for inverse function is
1. function must be injection e.g., one - one function
2. function should be surjective e.g., onto function ( codomain= range )
here, given,
f(x) = 2x - 3 ,
it is inversible function because it has both characters injective as well as surjective .
Let see how,
domain , co-domain and range of f(x) belongs to all real numbers.
Also this is one - one function ,
Because if we differentiate it ,
e.g., Dy/dx = 2 > 0 , we know, if f'(x) > 0 then f(x) be increasing function . Means it is one - one function [ every strictly increasing or decreasing Function is one - one function ]
Hence, function is one - one
Also, it is surjective , e.g., co-domain= range ∈R
Hence, f(x) = 2x - 3 is inversible function.
So, do inverse of it ,
y = 2x - 3
⇒ x = (y + 3)/2
⇒f⁻¹(x) = (x + 3)/2
I mean, conditions for inverse function is
1. function must be injection e.g., one - one function
2. function should be surjective e.g., onto function ( codomain= range )
here, given,
f(x) = 2x - 3 ,
it is inversible function because it has both characters injective as well as surjective .
Let see how,
domain , co-domain and range of f(x) belongs to all real numbers.
Also this is one - one function ,
Because if we differentiate it ,
e.g., Dy/dx = 2 > 0 , we know, if f'(x) > 0 then f(x) be increasing function . Means it is one - one function [ every strictly increasing or decreasing Function is one - one function ]
Hence, function is one - one
Also, it is surjective , e.g., co-domain= range ∈R
Hence, f(x) = 2x - 3 is inversible function.
So, do inverse of it ,
y = 2x - 3
⇒ x = (y + 3)/2
⇒f⁻¹(x) = (x + 3)/2
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