Consider the function y = 9 - x2, where x ≥ 3. What is the inverse of the function? What is the domain of the inverse? Show all of your work for full credit.

Answer:  [tex]y^{-1}=\sqrt{9-x}\qquad where\ x\leq 9[/tex]Step-by-step explanation:Given that y = 9 - x² where x ≥ 3 ⇒ y ≥ 0To find the inverse, swap the x's and y's and solve for y inverse (y⁻¹)[tex]y = 9 - x^2\\\\x = 9 - (y^{-1})^2\\\\(y^{-1})^2 = 9 - x\\\\\sqrt{(y^{-1})^2} = \sqrt{9 - x}\\\\y^{-1}=\pm\sqrt{9-x}\qquad \text{Since y}\geq 0, \text{then the negative is not valid}\\\\y^{-1}=\sqrt{9-x}[/tex]The domain is the restriction on the x-values of the equation.  Since the radicand (term under the square root symbol) must be greater than or equal to 0, then9 - x ≥ 09      ≥ x       ⇒     x ≤ 9