in set notation?
CORRECTION
I think the correct punctuation is √(8-x^2)
square root goes over all of it
y = √(8-x^2)
in order to find domain, we are going to equal it to zero as the following:
√(8-x^2) = 0
8 - x^2 = 0
8 = x^2
+/- 2√(2) = x
domain: [ -2√(2) , 2√(2) ]
Range, will be as:
when x = +/- 2√(2)
y = √(8 - ( 2√(2) )^2 ) = 0
when x = 0 because this is max where it reaches,
y = √(8 - 0^2) = +/-2√(2)
Range: [ 0 , 2√(2) ] <----- i took the positive because take a look at the graph at the following link:
http://www.wolframalpha.com/input/?i=y+%3D+%E2%88%...
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Domain = {x I - sqrt(8) ⤠x ⤠sqrt(8)}
range = {y I - sqrt(8) ⤠y ⤠sqrt(8)}
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Verified answer
y = √(8-x^2)
in order to find domain, we are going to equal it to zero as the following:
√(8-x^2) = 0
8 - x^2 = 0
8 = x^2
+/- 2√(2) = x
domain: [ -2√(2) , 2√(2) ]
Range, will be as:
when x = +/- 2√(2)
y = √(8 - ( 2√(2) )^2 ) = 0
when x = 0 because this is max where it reaches,
y = √(8 - 0^2) = +/-2√(2)
Range: [ 0 , 2√(2) ] <----- i took the positive because take a look at the graph at the following link:
http://www.wolframalpha.com/input/?i=y+%3D+%E2%88%...
=========
free to e-mail if have a question
Domain = {x I - sqrt(8) ⤠x ⤠sqrt(8)}
range = {y I - sqrt(8) ⤠y ⤠sqrt(8)}