#Squareroot as a function

14 messages · Page 1 of 1 (latest)

sly peakBOT
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worldly breach
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cause that's the outside function

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$u:=6+t^2$ makes $\sqrt{6+t^2}=\sqrt{u}$

marble oxideBOT
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Omegabet_

distant mulch
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Sqrt in itself isn’t a function

worldly breach
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it's a function with natural domain $[0,\infty)$

marble oxideBOT
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Omegabet_

distant mulch
worldly breach
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what that question has to do with whether a map is a function or not... no clue lol

distant mulch