How can a “proper” function have a vertical slope?

Plotting the function $$f(x)=x^{1/3}$$ defined for any real number $$x$$ gives us: Since $$f$$ is a function, for any given $$x$$ value it maps to a single y value (and not more than one $$y$$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $$x=0$$.

My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $$x$$. How is it possible for a "proper" function to have a vertical tangent?

As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?

• Funny that you mention this, because I recall, back in the day, being very upset with $f'(x)$ being the slope of the tangent line, since in most cases this so-called "tangent line" used to intersect the curve plenty of times, as opposed to just $1$. – Gae. S. Oct 5 at 18:08
• "In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function?" What do you think the slope of the tangent line at that point should be? – Noah Schweber Oct 5 at 18:25
• Rotate the graph through 90 degrees, and consider $f(x) = x^3$. Do you have a problem with the idea that the tangent line at $x = 0$ is horizontal? – alephzero Oct 6 at 14:21
• " To get a vertical tangent we need 2 vertical points". This certainly NOT true. I don't know where you tot that idea. That is true or a vertical graph, not of a vertical tangent to the graph. – user247327 Oct 6 at 15:20
• A vertical secant requires two points on the same vertical line... – Eric Towers Oct 6 at 16:43

No, we don't need two vertical points. By the same idea, if the graph of a function $$f$$ has an horizontal tangent line somewhere, that there has to be two points of the graph of $$f$$ with the same $$y$$ coordinate. However, the tangent at $$0$$ of $$x\mapsto x^3$$ (note that this is not the function that you mentioned) is horizontal, in spite of the fact that no two points of its graph have the same $$y$$ coordinate.

• +1 - verticality is a total red herring here! – Noah Schweber Oct 5 at 18:25
• @SolomonUcko wrong function. OP is talking about $x^{1/3}$, this answer says $x^3$. – jaxad0127 Oct 7 at 5:53
• Good catch @jaxad0127. Then I find the answer confusing. I suggest to change the answer to mention $x^5$ – Grzegorz Oledzki Oct 7 at 7:16
• @GrzegorzOledzki Instead of that, I've added a few words stating that my function is not the one mentioned in the question. – José Carlos Santos Oct 7 at 7:31
• Or just think of a circle - a tangent on a circle always only touches the circle in a single point. Although it describes the "direction" at this point, there are no two points of the circle on this line. – Falco Oct 7 at 13:11

The tangent line is simply an ideal picture of what you would expect to see if you zoom in around the point.

$$\hspace{8em}$$ That being said, the vertical tangent line to the graph $$y = \sqrt{x}$$ at $$(0,0)$$ says nothing more than that the graph would look steeper ans steeper as we zoom in further around $$(0, 0)$$.

We can also learn several things from this geometric intuition.

1. The line is never required to pass through two distinct points, as the idea of tangent line itself does not impose such extraneous condition.

For instance, tangent lines pass through a single point even in many classical examples such as conic sections. On the other extreme, a tangent line can pass through infinitely many points of the original curve as well.

2. Tangent line is purely a geometric notion, hence it should not depend on the coordinates system.

On the contrary, identifying the curve as the graph of some function $$f$$ and differentiating it does depend on the coordinates system. In particular, it is not essential for $$f$$ to be differentiable in order to discuss a tangent line to the graph $$y = f(x)$$, although it is a sufficient condition.

OP's example serves a perfect showcase of this. Differentiating the function $$f(x) = \sqrt{x}$$ fails to detect the tangent line at $$(0,0)$$, since it is not differentiable at this point. On the other hand, it perfectly makes sense to discuss vertical tangent line of the curve

$$\mathcal{C} = \{(x, \sqrt{x}) :x \in \mathbb{R} \} = \{(y^3, y) : y \in \mathbb{R} \},$$

and indeed the line $$x = 0$$ is the tangent line to $$\mathcal{C}$$ at $$(0, 0)$$.

• Woah! How did you make that amazing animation? – goblin Oct 6 at 6:14
• @goblin, Hi, I used Mathematica to make this animation. :) – Sangchul Lee Oct 6 at 19:26
• @SangchulLee, thanks! I can't wait to use Mathematica animations to improve my teaching. This is going to help me a lot. – goblin Oct 7 at 3:21

To get a vertical tangent we need 2 vertical points...

Herein lies the error in your assumptions. A tangent intersects a curve at the point of tangency in just one point.

• A tangent line may intersect the graph at different points. An extreme example is when the graph itself is already a line, say $y = ax + b$. Then the tangent line at any point coincides the graph itself, hence uncountably many intersections. A less extreme example would be $y = \cos(x)$. The tangent line to this graph at $(0,1)$ is exactly $y = 1$, and it intersects the graph at every $(2\pi k, 1)$ for $k \in\mathbb{Z}$. – Sangchul Lee Oct 5 at 19:43
• @SangchulLee I meant at the point of tangency. I'll include that now. Thank you for alerting me to this imprecision in language. – Allawonder Oct 5 at 19:45
• I believe the more precise version of this statement is that there is a neighborhood of the point of tangency that contains only one intersection. (Of course, this is not true for all curves, eg a line) You could mention something along those lines in the answer if you want. – David Z Oct 6 at 3:50
• @DavidZ I partly agree with you. However, to naturally include such degenerate cases as a line that's already right, I'll keep my statement about points of tangency. I thank you, though. – Allawonder Oct 6 at 5:23
• @DavidZ - not quite true. For example, the graph of $x^2\sin(1/x)$ at $x = 0$. – Paul Sinclair Oct 7 at 16:23

Vertical tangent does not mean that the function is not one to one.

The tangent line meets the function at point of tangency. The function is still one- to -one. If you find its inverse function you see the inverse is one to one and has derivative equal zero so your function is also one to one even if tangent line is vertical.

• This answer could potentially obfuscate to the asker what should be learned. I'll attempt to clarify, graph typically in set theoretical terms can be treated as synominous or the definition of "function". So replace graph with function. – marshal craft Oct 7 at 6:46
• @marshalcraft Thanks for the comment. I replaced graph with function to avoid misunderstanding. – Mohammad Riazi-Kermani Oct 7 at 10:07
• This answer is very correct I think, the asker basically associated tangent lines with vertical line test of functions which is wrong and unrelated. And you've very quickly gotten to that point. – marshal craft Oct 7 at 10:28

My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points...

As others have pointed out, this is the crux of the misunderstanding. That said, I'd like to try and succinctly highlight the core issue: and that is that derivatives are not defined by the secant of two points, but rather by the limit of secants approaching a certain point.

In the OP example, letting $$x = 0$$, as we take other domain values $$x'$$ trending closer to $$x$$, inspecting the trend of secant slopes is the essential meaning of the derivative there. In this case, the secant slopes grow larger without bound, which is the indication of an undefined (or infinite) derivative, and hence a vertical asymptote. The definition from Wikipedia:

Limit definition

A function $$f$$ has a vertical tangent at $$x = a$$ if the difference quotient used to define the derivative has infinite limit: $$\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = {+\infty}\quad\text{or}\quad\lim_{h\to 0}\frac{f(a+h) - f(a)}{h} = {-\infty}.$$

The Wikipedia article on vertical tangents uses the same $$f(x) = x^{1/3}$$ example, so hopefully that's clarifying. I suspect that this kind of misunderstanding may be the result of a particular course of study failing to emphasize the limit definition of the derivative.

• Good answer, I do think you have captured something as why this question was asked in first place. They are oblivious to limits. But most of the times limits aren't learned easily (some teachers even try to teach calculus without them, but I'm strong proponent of epsilon delta definition) so I don't know that this answer will find use. – marshal craft Oct 7 at 6:49

The "slope" at a single point of a function is really a limit of the slopes between points (this is the definition of the derivative, by the way). As you say, a "proper" function cannot have vertical slope as measured between two different points. To resolve the paradox, it pays to remember that taking limits can often result in extra "degenerate" cases to consider that one did not start out with. For example, think of a triangle which in the limit as two vertices are brought close to one another, becomes a line segment. And in the same manner here, functions can have points where the slopes near a point grow more and more vertical - and in the limit, we obtain a vertical tangent line.

By the way, your question is similar to some of the Zeno paradoxes, whose resolution eventually led to the concept of a limit. For a more "modern" take, one can give a more abstract answer behind this principle using topology, specifically the notion of boundary in topology. Oftentimes the class of objects we start out with is "open", meaning that a small perturbation of the object remains in the class (think of the class of triangles with positive area - we can move each vertex a little bit without causing the triangle to degenerate). But when we take limits, we can travel out of our open set and into its boundary. For the problem you asked about, the open class is the set of real numbers $$\mathbb R$$ representing the possible slopes of secant lines, and the degenerate case belongs to the extended real line, $$\mathbb R\cup \{\infty\}$$. Here the topological boundary is the singleton set $$\{\infty\}$$, representing the limiting case of a vertical - i.e., infinite - slope.

As you observed, $$\frac{dy}{dx}$$ is not defined at $$(x,y) = (0,0).$$ And there is no straight segment anywhere on the graph where the slope could be exactly measured by taking any two distinct points on that segment.

But those are not the only ways to define the slope of a curve.

One approach is to take a secant line between $$(0,0)$$ and a nearby point, and find the limit of the slope of that line as the distance goes to zero. You can also take the angle between the $$x$$ axis and the secant line between two points on either side of $$(0,0),$$ taking a limit as the distance between those points goes to zero. If the limit is defined (it is not for every curve) then it gives you the tangent line.

Another approach is to parameterize the curve: $$x = u(t)$$ and $$y = v(t)$$ where we let $$t$$ range over some (possibly infinite) interval of $$\mathbb R.$$ Consider the vector $$\left(\frac{d}{dt}u(t), \frac{d}{dt}v(t)\right).$$ As long as that vector is not $$(0,0)$$ we can consider it to be in the same direction as the slope of the curve. Now let $$x = u(t) = t^3,$$ $$y = v(t) = t.$$ Then the vector along the curve at $$(x,y) = (0,0)$$ is $$(0,1),$$ which points straight up, so the tangent line is vertical.

• I understand the first part of what you wrote, and perhaps my question is poorly phrased. so let me elaborate. say one secant point is at (0,0) and I another point is moving closer and closer to it (from either side, doesn't matter). How could we ever create a vertical line in this case (as the two secant points are never in a vertical line)? – Mike Oct 5 at 20:05
• It's the same principle as how you take the derivative at other points on the curve. None of the individual secant lines is vertical, but the angle of the line with the $x$ axis goes to $\frac\pi2$ ($90$ degrees) in the limit. – David K Oct 5 at 20:36

Let's find the tangent line equation of $$y=x^{1/3}$$ at the point $$x_0=\varepsilon>0$$: $$y=y(x_0)+y'(x_0)(x-x_0) \Rightarrow \\ y=\varepsilon^{1/3} +\frac{1}{3\varepsilon^{2/3}}(x-\varepsilon) \Rightarrow \\ x=-2\varepsilon+3\varepsilon^{2/3}y$$ Now for $$\varepsilon \to 0^+$$, we get the tangent line $$x=0$$. Similarly, we can show the left limit. At the $$x=0$$, the value of both curve and the tangent line $$y=0$$. Hence, the tangent line is $$x=0$$.

• Why the downvote? – farruhota Oct 6 at 21:23
• I don't know it your proof is correct, but maybe wouldn't be useful to askers. Also I did not downvote. – marshal craft Oct 7 at 6:51
• @marshalcraft, thank you for commenting. It answers the asker's question, so it must be useful. – farruhota Oct 7 at 7:16
• But if the asker does not have any notion of calculus (and probably at least a semester in real analysis) it is probably very unlikely that this will sink in, let alone answer anything for them. I've had many a professor that turn not to difficult topics, into incomprehensible nightmare. But like I said I didn't downvote but I don't think it is necessarily wrong to downvote. If it's correct then sure let it stay, but it serves no real use. – marshal craft Oct 7 at 10:04
• No real use to the asker* who is simply trying to maybe learn definition of function that we all learn at some point. And I guess one common pitfall of that test is near vertical sections? – marshal craft Oct 7 at 10:24

My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $$x$$. How is it possible for a "proper" function to have a vertical tangent?

This happens usually when the point of investigation (grey dot) is a point of inflection where the concavity of graph changes plus it is also a point where the tangent is parallel to $$y-$$axis. Since both the events taking place at a unique point $$x=0$$ of the domain, so it still qualifies the criteria to be a function (vertical line test). The function $$y=x^3$$ also has the same property of concavity change at $$x=0$$ but the tangent there is parallel to $$x-$$axis.

Just another point here, the function is a proper function, because it never really goes vertical. But the rate of change in values (derivative) when viewed as a function is not. To make this clearer that these are two different functions, you can again take the rate of change of this second function and view it as a function, so on and so forth. You keep getting different new functions forever that aren't the original, in this case.

Alternatively you can say the exact same thing but using the language of calculus and say the function passes the vertical line test but it's derivative does not. Taking a second derivative gives the acceleration. This function is infinitely many times differentiable so lies in $$C$$. Also this function does not equal any of its subsequent derivative.

So you are correct that no two points are collinear and vertical. But you are fundamentally incorrect in the assumption that a tangent line must share two points. See the tangent line of a circle, it oscilesses the edge of the circle or kisses it. Though tangent lines aren't necessary and your incorrect assumption about them isn't even necessary to see what you ask.

• Yeah again tangent lines aren't necessary, or related and that IS your mistake. You've rushed over definition of vertical line test. A vertical line collinear to at least 2 points in the range of the function. Instead youve used a vertical tangent line to the function which is incorrect. – marshal craft Oct 7 at 10:09

The derivative of the function is

$$y'=\frac{1}{3x^{2/3}}$$

$$y'(0)$$ is undefined due to division by zero, but

$$lim_{x \to 0} \,\, y' = \pm\infty$$

Alternatively, $$x$$ can be expressed as a function of $$y$$.

\begin{aligned} x &= y^3 \\ x' &= 3y^2 \\ x'(0) &= 0 \\ \end{aligned}

Since $$x'(0) = 0$$, the original function ($$y = x^{1/3}$$) must have a vertical tangent at $$x=0$$, because $$x$$ does not change when $$y$$ changes around that point.

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