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!
