The function \( g(x)=(x-3)^{2}+6 \) for \( x \leq 3 \), is one-to-one. a. Write an equation for the inverse function. b. State the domain and range of \( g \) and \( g^{-1} \).

To find the inverse function of \( g(x) = (x-3)^2 + 6 \) for \( x \leq 3 \), we will follow these steps: ### a. Finding the Inverse Function 1. **Set \( y = g(x) \)**: \[ y = (x-3)^2 + 6 \] 2. **Solve for \( x \)** in terms of \( y \): \[ y - 6 = (x-3)^2 \] \[ \sqrt{y - 6} = |x - 3| \] Since \( x \leq 3 \), we have \( x - 3 \leq 0 \), so: \[ \sqrt{y - 6} = -(x - 3) \] This simplifies to: \[ x - 3 = -\sqrt{y - 6} \] \[ x = 3 - \sqrt{y - 6} \] 3. **Express the inverse function**: \[ g^{-1}(y) = 3 - \sqrt{y - 6} \] 4. **Replace \( y \) with \( x \)** to write the inverse function: \[ g^{-1}(x) = 3 - \sqrt{x - 6} \] ### b. Domain and Range 1. **Domain and Range of \( g(x) \)**: - The function \( g(x) = (x-3)^2 + 6 \) is a parabola that opens upwards, with its vertex at \( (3, 6) \). - Since \( x \leq 3 \), the minimum value of \( g(x) \) occurs at \( x = 3 \), which gives \( g(3) = 6 \). - As \( x \) decreases from 3, \( g(x) \) increases without bound. - Therefore, the domain of \( g \) is \( (-\infty, 3] \) and the range is \( [6, \infty) \). 2. **Domain and Range of \( g^{-1}(x) \)**: - The inverse function \( g^{-1}(x) = 3 - \sqrt{x - 6} \) is defined for \( x - 6 \geq 0 \), which means \( x \geq 6 \). - As \( x \) increases from 6, \( g^{-1}(x) \) decreases without bound (since \( \sqrt{x - 6} \) increases). - Therefore, the domain of \( g^{-1} \) is \( [6, \infty) \) and the range is \( (-\infty, 3] \). ### Summary - The inverse function is: \[ g^{-1}(x) = 3 - \sqrt{x - 6} \] - The domain and range are: - For \( g(x) \): Domain \( (-\infty, 3] \), Range \( [6, \infty) \) - For \( g^{-1}(x) \): Domain \( [6, \infty) \), Range \( (-\infty, 3] \)