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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.

MSKB
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  • Where did you find that definition? – José Carlos Santos Feb 18 '21 at 17:52
  • Heard from a youtube video.I might be wrong. But this is what I understood from that video. – MSKB Feb 18 '21 at 17:54
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    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. – lulu Feb 18 '21 at 17:54
  • 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 $f(x_0,y_0)=0$ – lulu Feb 18 '21 at 17:57
  • 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? – MSKB Feb 18 '21 at 17:58
  • Well...don't forget we are talking about a level set, $f(x,y)=0$ or something like that. In that situation, you clearly can not move $x,y$ independently. Your example, $f(x,y)=x^2+y^2=25$ is perfectly good. The fact that we can (sort of) solve for $y$ as a function of $x$ just illustrates the Theorem. In general, one doesn't expect to get sensible formulations for the explicit function. – lulu Feb 18 '21 at 18:01
  • Does the plus minus thing (I mean the two values of y in the equation of a circle) makes an implicit function different from explicit function? – MSKB Feb 18 '21 at 18:09
  • No. I suggest ignoring whatever they said on the video. One could, for instance, take the function $f(x,y)=x-y$. Then the level set $f(x,y)=0$ just defines the function $y=x$. Perfectly good illustration of the Implicit Function Theorem. Again, it is certainly true that one doesn't expect to get sensible closed formulations of the implicitly defined function, even though in special cases you can. – lulu Feb 18 '21 at 18:12
  • Consider the implicit equation $y^x=x^y$. You can't isolate $x$ and $y$ using standard elementary functions. – PM 2Ring Feb 18 '21 at 18:16
  • @PM2Ring Yes that is different since we cannot express any of these variables independently,but in case of equation of circle we can express y=+-√(a^2-x^2). – MSKB Feb 18 '21 at 18:24

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One can find a rigorous definition in Wikipedia , 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.

Sometimes we can write implicit functions as a 'multiple' explicit function, for example we see that when writing the circle's equation:

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

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.

john
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Clemens Bartholdy
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