Vertical Tangent or Vertical Cusp?

[nasi112]

I have a concern about the function $f(x) = \sqrt{x}$. Does it have a vertical tangent or a vertical cusp at $x = 0$?

 

[fresh_42]

The derivative is not defined for $x=0,$ so there is no tangent.

However, if you look at the graph $\Gamma(f)=\{(x,f(x))\,|\,x\in \mathbb{R}\}$ as a differential manifold, the point $(0,f(0))$ is its boundary, and you can define a one-sided differential. But this is a very special case and point of view. From an analytical point of view, there is no tangent since we don't have an open neighborhood around $x=0$ in which the function is defined. So you have to decide whether you want to perform calculus or differential geometry.

 

[nasi112]

If I have to decide, I will choose calculus. The derivative is not defined at x = 0 for the function $f(x) = \sqrt[3]{x}$, still it has a vertical tangent.

 

[fresh_42]

If I have to decide, I will choose calculus. The derivative is not defined at x = 0 for the function $f(x) = \sqrt[3]{x}$, still it has a vertical tangent.

That depends on what you call a tangent. Let's take the function $f(x) = \sqrt[3]{|x|}$ just to avoid nasty complex roots and have it defined everywhere. I learned that the tangent is the limit of secants where the two intersection points converge to each other. Then the secants built from the points $(x,f(x))$ and $(0,0)$ get you a vertical line at the origin, but the secants between the points $(-x,f(-x))$ and $(x,f(x))$ result in a horizontal line. Which one should it be? Why is one method superior to the other one?

A tangent need differentiability in my understanding, and neither $f(x) = \sqrt[3]{x}$ nor $f(x) = \sqrt[3]{|x|}$ is differnetiable at the origin. Only if we pretend there is nothing other than our function graph (differential geometry) will we get a vertical line at the left boundary. If we have a neighborhood around the origin, then we don't have a tangent due to the lack of uniqueness.

 

[nasi112]

Vertical tangent - Wikipedia

en.wikipedia.org

This page was given previously in Najat's thread. It explained the difference between vertical tangents and vertical cusps pretty well. Unfortunately it didn't say anything about functions that has points with only one sided limit such as the derivative of $f(x) = \sqrt{x}$ at x = 0. Can I use this page to decide if the function $f(x) = \sqrt{x}$ has a vertical tangent, a vertical cusp, or neither?

 

[fresh_42]

Wikipedia has to be taken carefully. As I already said, it depends on how you define a tangent. It is the world of calculus versus the world of (differential) geometry. The Wikipedia entry is a bit sloppy since it mixes both perspectives. I could go into more detail and talk about local coordinates and differentiability at the boundaries, but it all comes down to your definition of a tangent and whether it is well-defined.

Infinity isn't a number and should never be treated as such, although things like L'Hôpital's rule allow us to operate with it.

If we have only the curve, then we can rotate it without changing anything and get a tangent. If we are stuck in the entire plane, then there is no tangent since it cannot be uniquely defined, and infinity isn't a number, and therefore no slope. Wikipedia mixes these two concepts.

 

[nasi112]

Let's forget for a moment the tangent is the limit of secants where the two intersection points converge to each other. Let's also assume the Wikipedia entry is not sloppy. How to use the limit definition of the vertical tangent written in that Wikipedia's page on the function $f(x) = \sqrt{x}$ at x = 0?

 

[fresh_42]

I don't know how to pretend that sloppy isn't sloppy. Even the Wikipedia article says $m_c$ undefined, and that is the problem. The derivative is not defined there, so we cannot take it to define the slope of a line. The tangent line is given as $$y=f(0)+f'(0)\cdot (x-0).$$ This means for our situation that
$$y=0+\infty \cdot x,$$and this cannot be turned into something looking like $x=a.$ I can't believe I have written infinity this way. That Wikipedia page should be deleted without replacement. We cannot even rotate the function since it stops being a function then. All we have is the graph.

You can patch two functions to get an example:
$$\begin{array}{lll} f(x)=\begin{cases} \sqrt[3]{x} &\text{ if } x\geq 0\\-\sqrt[3]{|x|}&\text{ if } x\lt 0\end{cases} \end{array}$$
Now you have the situation of the one on the Wikipedia page, however, $f$ is still not differentiable at $x=0.$ What we really have is an asymptotic behavior of $f',$ and you want to call the asymptote a tangent.

 

[nasi112]

I know in the real sense the tangent can't be construced on a point that is not differentiable. I know it's not a real tangent and it's only called tangent. I'm just saying let's pretend we are blind and apply their definition.

$$f(x) = \sqrt{x}$$
$$\displaystyle f'(x) = \frac{1}{2\sqrt{x}}$$
$$\lim_{x \to 0^{+}} \frac{1}{2\sqrt{x}} = \infty$$
I can't calculate the limit from the left side because the function's domain is $x \geq 0$. In this case, what is the conclusion according to their definition? Is there at x = 0, a vertical tangent, a vertical cusp, or neither?

 

[fresh_42]

According to their definition, it is a vertical tangent, no cusp, because the cusp requires two-sided, and different infinite limits.

What actually happens is that if you approach zero from the right, then the tangents become steeper and steeper. But one cannot reach infinity. Your tangent line will be hanging at the point zero and swinging around at any value of slope if you reach the boundary. Differentiability and continuity, too, are local properties. That means they have to hold in a neighborhood around the point of investigation, and here we don't have anything left of the point, no neighborhood. The question is what we do instead. Vertical tangents are such a workaround, but I wouldn't use them.

 

[nasi112]

Thanks. This is what I wanted to achieve. Now what happens if we extend the function?

$$f(x) = \sqrt{|x|}$$
How to take the derivative of this function?

 

[fresh_42]

I would use the chain rule. If $g(y)=y^{1/2}$ and $h(x)=|x|$ then $f=g\circ h.$ The chain rule then gives us $f'(x)=g'\cdot h'.$ The derivative of $g$ is $g'=(1/2)|x|^{-1/2}$ and $h'=\operatorname{sgn}(x),$ the sign of $x.$ Thus
$$f'(x)=-(1/2)\operatorname{sgn}(x)|x|^{-1/2}=\begin{cases}-\dfrac{1}{2\sqrt{|x|}}&\text{ if }x<0\\[10pt] \dfrac{1}{2\sqrt{|x|}}&\text{ if }x\ge 0\end{cases}$$We can always write $\operatorname{sgn}(x)=\dfrac{x}{|x|}$ which means $f(x)=\dfrac{x}{2\sqrt{|x|^3}}.$ The function is not differentiable at $x=0$ and has a cusp there.

 

[nasi112]

The derivative is a little complicated but I got it. The cusp happens in the extended function because the two-sided limits don't match, right?

 

[fresh_42]

The derivative is a little complicated but I got it. The cusp happens in the extended function because the two-sided limits don't match, right?

Yes, plus they are infinitely large. Would be interesting to know if Wikipedia also calls it a cusp if e.g. the limits were $\pm 3$ instead of vertical.

Edit: The last $f$ in post #12 should of course be $f'$ but it's too late to edit.

 

[nasi112]

Yes, plus they are infinitely large. Would be interesting to know if Wikipedia also calls it a cusp if e.g. the limits were $\pm 3$ instead of vertical.

Edit: The last $f$ in post #12 should of course be $f'$ but it's too late to edit.

Thanks. I got it.