Consider f:R+→[0,∞) given by f(x)=5x²+6x+0. Prove that f is invertible with f⁻¹(y)=(√(54+5y)-3)/5.

The domain is restricted to positive real numbers only, so:
y=f(x)=5x²+6x+0=5((x+3/5)²-9/25) ⇒(x+3/5)²=9/25+y/5 ⇒x=(-3/5 ± √(9/25+y/5))
Since x must be positive, we take the positive root:
x = (√(9/25+y/5)-3/5) = (√(9+5y)/5 - 3/5) = (√(9+5y)-3)/5
for all y in the range.
Thus the function is onto.
Now let's assume x₁≠x₂, then y₁=y₂ ⇒ 5((x₁+3/5)²-9/25) = 5((x₂+3/5)²-9/25) ⇒ x₁ = x₂, which is a contradiction to our assumption and therefore x₁=x₂.
Thus the function is one-one.
Since the function is both one-one and onto, it is invertible on the given domain. Its inverse is given by the expression for x found above while proving that the function is onto.