Question

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 1

Answer:
`y^(-1)=√(9-x)\qquad where\ x\leq 9`

Explanation:

Given that y = 9 - x² where x ≥ 3 ⇒ y ≥ 0

To find the inverse, swap the x's and y's and solve for y inverse (y⁻¹)

`y = 9 - x^2\\\\x = 9 - (y^(-1))^2\\\\(y^(-1))^2 = 9 - x\\\\\sqrt{(y^(-1))^2} = √(9 - x)\\\\y^(-1)=\pm√(9-x)\qquad \text{Since y}\geq 0, \text{then the negative is not valid}\\\\y^(-1)=√(9-x)`

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, then

9 - x ≥ 0

9 ≥ x ⇒ x ≤ 9