How to know which line represents tangent to a curve $y=f(x)$ (in RED) ?From the diagram , I cannot decide which line to take as tangent , all seem to touch at a single point.

25It does not exist for the reason you give. The tangent, if it exists, is unique. – John Douma Mar 18 '19 at 16:06

2how about the bisector of the "angle" made of the right and left part of your nice sharp curve? – dmtri Mar 18 '19 at 16:14

If the curve consists of 2 arcs, let's say parts of the circles with radius 1 and centers 1 and 1 then the axis $x=0$ might be the tangent . – dmtri Mar 18 '19 at 17:22

@dmtri because the "angle" might not exist. Consider a function like $y = 2x + x \sin(1/x)$, at $x = 0$. If you draw a graph, it "sort of" looks like it has a cusp at $x = 0$, but the tangent direction is "infinitely wiggly" on either side of the cusp. Trying to give a useful *mathematical* definition of what you mean by "a nice sharp curve" is hard! – alephzero Mar 18 '19 at 19:20

5A tangent is a line that approximates the curve near the point  such that if you zoom in close enough on the point, the tangent and curve become pratically indistinguishable. That is the property of tangents that makes them useful. Left and Right tangents, which approximate the curve on only one side of the point, are also useful concepts. A line that does not approximate the curve on either side is useful only in very rare and limited circumstances, unworthy of a special name. – Paul Sinclair Mar 19 '19 at 03:11

1@alephzero, I am not saying that such an angle should exist.The sketch of the OP is just a sketch and not mathematicaly described, so no strict results can be deduced. In my first comment I just wonder what the tangent would be if $f$ were something like $y=x^{2/3}+c$ – dmtri Mar 19 '19 at 08:03

You can think of a formal definition of "sharp point" as a point in which does not exist a unique tangent. That is just the nature of these cusps. – M. Winter Mar 20 '19 at 09:03

The tangent may or may not exist at that point, depending on whether or not the two "sides" are mutually tangent. If the left and right tangents coincide, the tangent *does* exist there. Tangent doesn't just mean "meeting at one point". The normal does that too, but it isn't a tangent. You could slide your star of lines to any point on the curve and make the same argument. – MPW Mar 20 '19 at 15:59
7 Answers
None of them, because it's not differentiable there, so there is no tangent line at that point.
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If I have suppose a plot $(y^24x)(y^24x)=0$ , I cannot differentiate this for all values of x? – Aditya Prakash Mar 18 '19 at 16:16

@ADITYAPRAKASH $(y^24x)(y^24x)=0$ is not a function, but a relation. How/where would you like to differentiate it? – Botond Mar 18 '19 at 16:23

I put the wrong one , sorry . I wanted the graph of $y^2=4x and y^2=4x$ together as a joint equation , positive parts only and to differentiate at x=0 – Aditya Prakash Mar 18 '19 at 16:27

6@ADITYAPRAKASH That's a relation too. But, for example, if you define $f$ as $f(x)=2\sqrt{x}$, then $f$ is differentiable for all $x \neq 0$. At $x=0$, the limit $\lim_{x \to 0} \frac{\sqrt{x}}{x}$ does not exists because when $x>0$, $\frac{\sqrt{x}}{x}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}} \to +\infty$ and when $x < 0$, $\frac{\sqrt{x}}{x}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}} \to \infty$, so the limit cannot exist. – Botond Mar 18 '19 at 16:33

If $f$ were something like $f(x)=x^{2/3}$, then there is no tangent line at $x=0$ ? – dmtri Mar 19 '19 at 08:06

3@dmtri How do you define $x^{2/3}$? If the domain is $\mathbb{R}_{+} \cup \{0\}$, then $0$ is not in the interior of the domain, so you can't differentiate it there. If you define it on $\mathbb{R}$, then it won't be differentiable at $0$, so you won't have a tangent line there. – Botond Mar 19 '19 at 08:31

In the whole of $\Bbb{R}$, and as you said there is no derivative at $0$, but why not a tangent line there? There is a lot of discussion about that in wiki. – dmtri Mar 19 '19 at 08:54


When the derivative exists the usual way, but when it does not exist , I really do not know and I am also confused about that... Maybe it debends on the domain you are studying.... In geometry might also exist 2 different tangent lines at a point of a algebraic curve.... – dmtri Mar 19 '19 at 09:31

@dmtri Honestly, I don't know how does it work in geometry. I'd define it as the degree $1$ taylor polinomial, i.e. if the function $f$ is differentiable at $a$ then it's tangent line at $a$ is $g(x)=f(a)+f'(a)(xa)$. But in geometry, you would define it for a circle as well, right? And it would have vertical tangent line "at the sides", right? But I think the tangent line is unique in geometry as well. – Botond Mar 19 '19 at 09:43

Please see also https://math.stackexchange.com/questions/1748162/problemwithbasicdefinitionofatangentline – dmtri Mar 19 '19 at 11:16
One of the defining points (no pun!) of a tangent is the idea that as you look at a smaller and smaller section of the curve, the curve starts to more and more looks like a straight line.
That behaviour lets you do a huge number of things:
 You can estimate where the curve goes in places near that point (if tangent has a gradient of 2, then the curve at X + 0.000001 will probably be close to Y+0.000002, or whatever).
 You can approximate the curve by tiny sections of straight lines  this is the basis of simple calculus.
 You can take limits in several ways, and expect them to converge as you use narrower sections of the curve
 ? Many other things, all stemming from these.
Not all curves behave like that, and yours doesn't. No matter how close you inspect the cusp of your curve (the pointy bit), it's never going to resemble anything like a straight line, on any scale. It'll always look like a cusp.
That's the fundamental reason there isn't a tangent at that point. Because the curve just doesn't resemble a straight line, even in a microscopic closeup view, it doesn't have a gradient or tangent at that point, and it isn't differentiable at that point (much the same thing at a very simple level), and so on.
That isn't in fact unusual. In fact more curves don't have a gradient than do  its just that we don't study curves at random, so you mainly look at curves that do  at least, at this level of maths!
Other examples of curves that don't have a gradient at some or all points 
 the "step" curve (defined as y=0 if x<0, y=1 if x >= 0)
 a "curve" defined as y=1 if x contains a "1" when written in decimal, and y=0 otherwise. Because there are infinitely many numbers that simply don't have a 1 in their decimal expansion, slotted in between those that do!
 the curve y = 1/x at x=0
 the "blancmange curve", which doesn't have any breaks in it, and looks like it should be a nice straightforward curve, but actually "wobbles" so much at any microscopic level (however closely you look at it,and wherever you look!) that in the end, it doesn't have a gradient anywhere.
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If the cuspid you draw is actually a cuspid (roughly speaking: if microscopically it is not "rounded") you can speak of a right and left tangent, same as you do with limits.
Refer to this wiki article.
Also note that, if the semiderivative on one side has the value of $m$ and the other $\,m$, then the two lines geometrically coincide into a single line (or, maybe better, half line or ray).
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6To this I might add that for one point, left derivative = right derivative (and they both exist) $\iff$ differentiable $\iff$ has a (unique) tangent. And also cite [wikipedia](https://en.wikipedia.org/wiki/Semidifferentiability) – asky Mar 18 '19 at 23:17


1@asky as the answer points out, unique tangent is equivalent to left derivative $=\pm$ right derivative rather than differentiable. – jgon Mar 19 '19 at 00:22

2@jgon that depends how you define tangents being equal. From an analytic geometer's perspective, they have opposite slopes, so are not equal, but from a classical geometer's perspective, they look the same, so are equal. I think there is no canonical equivalence relation on lines, and both views accept an equivalence relation so are equally valid. – asky Mar 19 '19 at 01:43

2@asky I suppose that's a reasonable viewpoint. I'm suppose I'm just much more used to thinking of lines as unoriented. – jgon Mar 19 '19 at 01:48
The "classical" tangent does not exist indeed. However for a convex function with a sharp point, you can use a generalization of the derivative called the subderivative. It is used frequently in optimization.
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3While I thikn mentioning the subderivative is definitely relevant here, the function of OP seems to be not convex at all. However, in this case you can take the generalized Clarke gradient (which reduces to subderivative when the function is convex). – Surb Mar 18 '19 at 20:20
If $y=F(x)$ is not differentiable then the differential coefficient /derivative does not exist and there will be no unique tangent defined.
Ignore the following if it makes no sense now. There are situations when if a third space variable $z$ like in $(x,y,z) $ is defined involving a fourth common parameter $t$ like it is for this four branched curve between four sharp cusps:
$$ x= \cos^3t, y= \sin^3t, z= \frac12 \cos^2t $$
At $ xz$ or $ yz$ projections ( a shadow, if you will) exist tangents that suddenly rotate by $90^{\circ}$ at cusp points and tangent slope includes all slopes in between.. in a situation where there is differentiability at these critical points.
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Since we're doing geometry, let's think kinematically, too. Imagine a point moving along your curve, in the direction of increasing $x.$ Furthermore, let a ray project from the point, tangent to the curve there (you might think of your curve as a lane, and the point a car in the night; imagine the headlights beaming straight ahead as the tangent ray at any location on the curve).
Then as you approach the cuspoid from the left, the ray points down; at that point therefore, we can say it points down from the left. Now, let the point turn without leaving the point; then the tangent ray points up to the right  at the same point! That happens nowhere else on the curve. Thus, we have two possible candidates for the tangent using this definition of tangent. If we require the tangent at a point to be unique, then there is no way to pick one of the two at the cuspoid without making further assumptions (which will be more or less arbitrary). The same thing happens for a point approaching the offending point from the right along the curve.
Thus, the curve has no unique tangent at that point. The reason is that you can think of the two branches of the curve as separate curves intersecting at a nonzero angle at that point. Suppose, for example, that we have two semicircular arcs instead, of the same radius and touching externally. For more clarity, take the positive parts of $(x+1)^2+y^2=1$ and $(x1)^2+y^2=1.$ Then if we do not consider out tangents as directed, we can have a unique tangent at the origin in this case, since the two arcs are parallel at that place. However, if we think in terms of functions, there is still no derivative at that point, since the derivative measures the slope of the tangent line wrt increasing $x.$ However, if you imagine our point moving along this curve from left to right and crossing the origin, then the tangent ray points in opposite directions immediately before and after the point. Thus, there is no derivative there. However, purely geometrically, it is possible to speak of a tangent in this case.
I hope I have contributed to your understanding.
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There are two possible definitions of "tangent" at a point $\vec{p}$ in terms of the equation $f(\vec{v})=0$ of the curve as follows.
Given a line $(1t)\vec{p}+t\vec{q}$ (where $\vec{p}$ and $\vec{q}$ are distinct points on the line) we substitute this in the equation to obtain a function $g(t)$. The multiplicity of intersection of the curve and the line at $\vec{p}$ is the order of vanishing of $g(t)$ at $t=0$ (which could be "infinite").
Definition 1 : A tangent line is a line which meets with multiplicity at least 2.
This is the most common definition. With this definition, there are infinitely many tangents at the point in your figure.
Definition 2 : A tangent line is a line which meets with the maximal multiplicity.
With this definition, there is a single such line in your example. However, we can easily make a figure with two or more cusps joined together and thus create an example with more than one tangent under this sense of tangent.
To avoid confusion, the second kind of tangent is usually not called a tangent but an "infinitely near point" in algebraic geometry. (However, this notion makes sense for more general situations than the case where $f$ is a algebraic.)
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