Consider f:R+→[4,∞] given by f(x)=x²+4. Show that f is invertible with the inverse f⁻¹ of f given by f⁻¹(y)=√y where R+ is the set of all non-negative real numbers.

f:R+→[4,∞] is given as f(x)=x²+4
For one-one:
Let f(x)=f(y)
x²+4=y²+4
x²=y²
x=y [as x=y∈R+]
Therefore, f is a one-one function.
For onto:
For y∈[4,∞], let y=x²+4
x²=y−4≥0 [as y≥4]
x=√y−4≥0
Therefore, for any y∈[4,∞] there exists x=√y−4∈R+, such that f(x)=f(√y−4)=(√y−4)²+4=y−4+4=y
Therefore, f is onto.
Thus, f is one-one and onto and therefore, f⁻¹ exists.
Let us define g:[4,∞]→R+ by g(y)=√y
Now,
(g∘f)(x)=g(f(x))=g(x²+4)=√(x²+4)−4=√x²=x
And
(f∘g)(y)=f(g(y))=f(√y)=(√y)²+4=y+4=y
Therefore, g∘f=f∘g=IR
Hence, f is invertible and the inverse of f is given by f⁻¹(y)=g(y)=√y