+0  
 
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\(y = {\sqrt{2-x^{2}}}\)

 

The answer is {y│0 ≤ y ≤ 2, y ∈ ℝ}, I'm completely lost!

 Oct 22, 2018
 #1
avatar+3593 
+2

\(\text{well the domain of y(x) is }[-2,2] \\ \text{and over this domain y increases from 0 to }\sqrt{2} \text{ and then decreases back to 0}\\ \text{so the range of y(x) is}\\ \left\{y:y \in [0,\sqrt{2}]\right\}\\ \text{what you've written isn't quite correct. y(x) never gets above }y=\sqrt{2}\)

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 Oct 22, 2018

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