We know, for a circle of radius 1 centred at the origin: $$ y^2 + x^2 = 1$$ Now when I saw the graph of the given equation- any x-value in within the domain has two images, but how can that be? Since I've read that in a function any pre-image cannot have two images.
And now the above can be equivalently written as, $$y = \sqrt{1-x^2} $$ But now, surprisingly(for me), any x-value doesn't have two images! But why is that?
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From the equation $$x^2+y^2=1$$ we can obtain the functions $$f(x) = \sqrt{1-x^2},$$ $$g(x) = - \sqrt{1-x^2},$$ whose union of their graphics is the unit circle.
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From $y^2=x$, the conclusion $y = \sqrt{x}$ doesn't necessarily follow. For instance, from $(-3)^2 = 9$, we cannot conclude that $-3 = \sqrt{9}$. That's because for $x \geq 0$, we define $\sqrt{x}$ as the unique positive number $y$ satisfying $y^2 = x.$
(The previous sentence uses 'positive' in a non-standard way. In particular, in this context I mean 'equal to $0$, or greater than zero.' We really should come up with a new adjective for this condition.)
This means that the square root function doesn't necessarily undo the squaring function. For example, from $(-3)^2 = 9$, we can correctly deduce that $\sqrt{(-3)^2} = \sqrt{9}$. But it's incorrect to simplify the left-hand-side to $-3$, thereby concluding $\sqrt{9} = -3$. Indeed, the expression $\sqrt{(-3)^2}$ turns out to equal positive three, not negative three. We can summarize this as:
Common False Belief. For all real numbers $x$, we have $\sqrt{x^2} = x.$
(However, the above principle is true if we assume $x \geq 0$.)
In more detail, the reason for this subtlety is that any $x > 0$ has exactly two square roots (i.e. numbers $y$ satisfying $y^2=x$) in the real line, and they differ by a factor of $-1$. For example, the square roots of $9$ are $3$ and $-3$. So we define $\sqrt{x}$ to be the positive one, and obtain the other square root as $-\sqrt{x}.$ For instance, $\sqrt{9} =3$, hence the other square root is $-\sqrt{9} = -3$.
Following this line of thought, which emphasizes that every $x > 0$ has precisely two square roots, that we write $\sqrt{x}$ for the positive one, and that we can write $-\sqrt{x}$ for the negative one, we obtain the following principle of mathematics:
Theorem. For all real numbers $x$ and $y$, we have:$$y^2 = x \;\;\leftrightarrow\;\; y = \sqrt{x} \vee y = -\sqrt{x}.$$
Using this, we can argue as follows.
Consider real numbers $x$ and $y$. Then the following are equivalent:
So basically, we've expressed written the unit circle as a union of the upper half and the lower half.
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No. For instance, it doesn't pass the vertical line test, which means there are two $y$-values for an $x $-value. ..
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Something is a function if, and only if, there is only one value of y for each value of x. That's the definition of function.
Why is so? Because the idea of function is that you can relate two values in a determinist way, so that you know the value of the dependent variable if you are given the value of the independent one. If however there are two (or more) different possible values for y for a given value of x, then y doesn't actually depend exclusively on x, ie, it isn't a "function of x".
Because there is also $$y=-\sqrt{1-x^2}.$$
The circle with equation $x^2+y^2=1$ it's not a graph of a function.
We can understand it by another algebraic way. $$\left(\frac{3}{5},\frac{4}{5}\right)$$ and $$\left(\frac{3}{5},-\frac{4}{5}\right)$$ placed on the graph and this is a contradiction with a definition of the function.