Question

f(x)= 15-x^2 x greater than or equal to 0
f^-1 (x)= ?

Answer 1

put
\[x=15-y^2\] and solve for y

Answer 2

you get
\[y^2=15-x\] and so
\[y=\pm\sqrt{15-x}\] but since the domain of your original function was positive numbers your answer is
\[f^{-1}(x)=\sqrt{15-x}\]

Answer 3

Got it! Thanks