Q. 115.0( 2 Votes )

Consider f : R<su

Answer :

A function is said to be invertible if it is one-one onto.


To prove that f(x) is invertible we need to prove that it is one-one onto.


As f(x) = x2 + 4


And domain is all non-negative real numbers.


No two values of x is going to give same results as x is a non-negative.


So for every unique value of x there is only one unique value of f(x)


Hence, we say that f(x) is one – one.


f(x) = x2 + 4 ≥ 4


range of f(x) = [4, ∞)


As, f : R+ [4, ∞) [given mapping of f]


Range = [4,∞) = co-domain


f(x) is onto


As, f(x) is one-one and onto.


f(x) is invertible.


Now, we need to find the inverse of the function f(x)


Let, y = x2 + 4


x2 = y – 4


x = ± √(y-4)


As domain of f is [0,∞) {given}


x > 0


x = √(y-4)


Or we can write: f-1(y) = √(y-4)


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