$x^2 + y^2= 36,\ y\in\mathbb [0,6] $ is a function or not ?
My attempt:
$x=\pm\sqrt{36-y^2}$
Since for every value of $y$, $x$ has two possible values, so the given relation is not a function .
But my book says that it is a function. Why?
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First thing is that it doesn't make sense to ask "is this equation a function?" If the convention is that the real question here is "is $y$ a function of $x$?" then the question makes sense.
That relation also stipulates that $y\in[0,6]$. The graph of this relation is just the upper half of a circle. Under that condition, the relation is equivalent to $y=\sqrt{36-x^2}$ and there is no $\pm$. Each value of $x$ (in $[-6,6]$) corresponds to exactly one value of $y$. So the relation here makes $y$ a function of $x$.
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then $ x^2 +y^2=36, x\in\mathbb[0,1]$ is a function also – Abhishek Kumar Apr 06 '19 at 02:57
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@AbhishekKumar No. If $x \in [0, 1]$, it is possible to obtain two $y$-values for each $x > 0$. For instance, if $x = 1$, $y = \pm \sqrt{35}$. – N. F. Taussig Apr 06 '19 at 09:14