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According to the definition of implicit function we cannot determine the value of one variable explicitly from the function. I have gone through many websites, few books and few youtube videos but could not understand implicit function clearly.

Equation of a circle is an implicit function as mentioned in most of the sources because we cannot get the value of y from the function f(x,y)=x^2+y^2=25 explicitly.But why can't we get the value of y explicitly i mean according to this function we can get two values for y at a certain abcissa. We are clearly getting two values of y for a certain value x then why is it implicit function?

I could not get the concept of this type of function. Pardon me for writing anything wrong if I have done any.Looking for an explanation

Again I apologise for asking such a silly question in this forum.

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  • $\begingroup$ Where did you find that definition? $\endgroup$ Feb 18, 2021 at 17:52
  • $\begingroup$ Heard from a youtube video.I might be wrong. But this is what I understood from that video. $\endgroup$
    – MSKB
    Feb 18, 2021 at 17:54
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    $\begingroup$ Well, an explicit function is one given explicitly, as in $y=f(x)$. By contrast, an implicit function is defined via $f(x,y)=$constant. It may, of course, be possible to solve that explicitly to get $y$ as a sensible function of $x$, but usually it is not. $\endgroup$
    – lulu
    Feb 18, 2021 at 17:54
  • $\begingroup$ The big thing to keep in mind is the Implicit Function Theorem which, loosely speaking, gives you reasonable conditions under which writing something like $f(x,y)=0$ guarantees the existence of some function $g$ with $y=g(x)$ defined near some solution $f(x_0,y_0)=0$ $\endgroup$
    – lulu
    Feb 18, 2021 at 17:57
  • $\begingroup$ Does that mean in y=f(x) y is a dependent variable that depends on the independent variable x and in case of f(x,y), x and y both are independent? $\endgroup$
    – MSKB
    Feb 18, 2021 at 17:58

1 Answer 1

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One can find a rigorous definition in [Wikipedia][1] , however I will attempt to provide a practical way of thinking of implicit function's.

An explicit single variable function can be thought of as a 'mapping' from one set to another set of the form $x \to y$, this can be graphed on a cartesian grid by highlighting all the pair $(x,y)$ included in this mapping.

Now, in an implicit function, we no longer have the idea of mapping from the previous definition but we can think of the solution's to the implicit equation which defines the function. If we took a graph and highlighted all the points which solved the equation of the implicit function, then we would get the curve of the implicit function analogous to how there was a curve corresponding to the explicit function.

Some times we can write implicit functions as an 'multipile' explicit function, for example how we see writing the circle's equation as:

$$ y = \begin{cases} \sqrt{1-x^2} \\ - \sqrt{1-x^2} \end{cases}$$

Notice that we have two possible definitions of 'y' when we isolate the implicit equation and solve for $y$, this two definitions can be thought of as pieces of the original curve. For example, $y=\sqrt{1-x^2}$ defines the upper disc and $ y = - \sqrt{1-x^2}$ defines the lower one.

[1]: https://en.wikipedia.org/wiki/Implicit_function#:~:text=An%20implicit%20function%20is%20a,)2%20%E2%88%92%201%20%3D%200

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  • $\begingroup$ ""Some times we can write implicit functions as an 'multipile' explicit function, for example how we see writing the circle's equation as: y={1−x2−−−−−√−1−x2−−−−−√"" $\endgroup$
    – MSKB
    Feb 18, 2021 at 18:40
  • $\begingroup$ This is the thing which I was looking for to be honest.....Thank you very much. $\endgroup$
    – MSKB
    Feb 18, 2021 at 18:41
  • $\begingroup$ You meant multivalued explicit function right? $\endgroup$
    – MSKB
    Feb 18, 2021 at 18:44
  • $\begingroup$ I'm not sure if you could use the term multivalued here, but more important that name is idea see here and linked for discussion of above $\endgroup$ Feb 18, 2021 at 21:48
  • $\begingroup$ Hmm on reading wiki, they do use multivalued implicit function, so that works ! see under algebraic functions $\endgroup$ Feb 18, 2021 at 21:56

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