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.
