For the function f (x) = square root (x-5), find f^-1. What is the range of f^-1? Any explanation and answer is appreciated!

F (x) = sqrt (x-5)

f^-1 (x) will be this ...

y=sqrt (x-5)

x=sqrt (y-5) (switch x and y)

x^2 = [sqrt (y-5) ]^2 (solve for y)

x^2=y-5

x^2+5=y

f^-1 (x) = x^2+5

The range of f^-1 (x) is the same as the domain of f (x) so ...

Find the domain of f (x)

x-5≥0

x≥5

interval notation: [5,∞)

That's the domain of f (x) and therefore it's the range of f^-1 (x)

If you want me to explain this further, tell me in the comments.

Best wishes!