solution:
Think about f : R+ → [4, ∞) given by f(x) = x2 + 4 Let x, y ∈ R → [4, ∞) then, at that point
f(x) = x2 + 4 and
f(y) = y2 + 4
on the off chance that f(x) = f(y) x2 + 4 = y2 + 4 or x = y
f is one-one.
Presently y = f(x) = x2 + 4 or x = √???? − 4 as x > 0 f(√???? − 4 )= (√???? − 4 )^2 + 4 = y
f(x) = y
f is onto work.
Consequently, f is invertible and Inverse of f will be f – 1 (y) = √???? − 4 .