\(y = {\sqrt{2-x^{2}}}\)
The answer is {y│0 ≤ y ≤ 2, y ∈ ℝ}, I'm completely lost!
+6248 \(\text{well the domain of y(x) is }[-2,2] \\ \text{and over this domain y increases from 0 to }\sqrt{2} \text{ and then decreases back to 0}\\ \text{so the range of y(x) is}\\ \left\{y:y \in [0,\sqrt{2}]\right\}\\ \text{what you've written isn't quite correct. y(x) never gets above }y=\sqrt{2}\)
.
