Step-by-step explanation:
[tex]f(x)=\sqrt{9-x^2}\to y=\sqrt{9-x^2}[/tex]
Restriction for x: 0 ≤ x ≤ 3.
Therefore the restriction for y:
[tex]f(0) = \sqrt{9-0^2}=\sqrt9=3\\\\f(3)=\sqrt{9-3^2}=\sqrt{9-9}=\sqrt0=0[/tex]
Restriction for y: 0 ≤ y ≤ 3
Exchange x to y and vice versa.
[tex]x=\sqrt{9-y^2}[/tex]
Solve for y:
[tex]x^2=\left(\sqrt{9-y^2}\right)^2\\\\x^2=9-y^2\qquad\text{add}\ y^2\ \text{to both sides}\\\\x^2+y^2=9\qquad\text{subtract}\ x^2\ \text{from both sides}\\\\y^2=9-x^2\to y=\sqrt{9-x^2}[/tex]
CONCLUSION:
[tex]f^{-1}(x)=f(x)[/tex]
The graph of a function and a function inverse to it are symmetrical about the line y = x
