help finding domain and range of inverse functions, please?

Question

equation: f(x) = x^2 + 4 x >= 0
Domain: (0, oo)
Range: (4,oo)

equation: f^-1(x) = ( x - 4) ^ (1/2)
Domain: (4, oo)
Range: (0, oo)

Is this correct? Thanks for your help.

Answer 1

Yes, these are correct. The range of the inverse function is the domain of the original function, and vice versa.

Answer 2

No, it's wrong. For the given domain, the range of f is (0, oo)

To finf f^-1, you solve tyhe equantion x^2 + 4x = y for y >0. By Bhaskara, you get 2 solutions x = -1 + (sqrt(4 + y))/2 and 1 - (sqrt(4 + y))/2. But since x is required to be in (0, oo), you must have f^-1(y) = 1 + (sqrt(4 + y))/2