1. **State the problem:** Find the domain and range of the function $$f(x)=(x-2)^2+3$$ with the restriction $$x \leq 2$$, and also find the domain and range of its inverse function. 2. **Domain of $$f$$:** Given the restriction $$x \leq 2$$, the domain is all real numbers less than or equal to 2, so $$\text{Domain}(f) = (-\infty, 2]$$. 3. **Range of $$f$$:** Since $$f(x)=(x-2)^2+3$$ is a parabola shifted right by 2 and up by 3, and $$x \leq 2$$ means we take the left half of the parabola, the minimum value occurs at $$x=2$$: $$f(2) = (2-2)^2 + 3 = 0 + 3 = 3$$ For $$x < 2$$, $$f(x) > 3$$ because the square term is positive. So the range is $$[3, \infty)$$. 4. **Find the inverse function:** To find $$f^{-1}(y)$$, start with $$y = (x-2)^2 + 3$$ Isolate the square term: $$y - 3 = (x-2)^2$$ Take the square root (considering the domain restriction $$x \leq 2$$, we take the negative root): $$x - 2 = -\sqrt{y - 3}$$ Solve for $$x$$: $$x = 2 - \sqrt{y - 3}$$ So, $$f^{-1}(y) = 2 - \sqrt{y - 3}$$ 5. **Domain of $$f^{-1}$$:** The expression under the square root must be non-negative: $$y - 3 \geq 0 \implies y \geq 3$$ So the domain of $$f^{-1}$$ is $$[3, \infty)$$. 6. **Range of $$f^{-1}$$:** Since $$f^{-1}(y) = 2 - \sqrt{y - 3}$$ and $$\sqrt{y - 3} \geq 0$$, the maximum value of $$f^{-1}$$ is at $$y=3$$: $$f^{-1}(3) = 2 - 0 = 2$$ As $$y \to \infty$$, $$f^{-1}(y) \to -\infty$$, so the range is $$(-\infty, 2]$$. **Final answers:** - $$\text{Domain}(f) = (-\infty, 2]$$ - $$\text{Range}(f) = [3, \infty)$$ - $$\text{Domain}(f^{-1}) = [3, \infty)$$ - $$\text{Range}(f^{-1}) = (-\infty, 2]$$