vertical tangent

Alan Najat

New member
Joined
Jun 11, 2024
Messages
9
hey, I've gotten really confused over something even though it is kinda simple I guess, once I saw a question that was asking if wether f(x)=x^(4/5) has a vertical tangent or not, I believed it had, but for certainty I asked Chat GPT and he said it has it, so I answered the question and said it has it, but a teacher that has master's degree in mathematics said no, it doesn't have a vertical tangent, it only has Cusp, I was confused, after several months I got back to the same problem but this time I checked the CALCULUS book as a source to know for sure what is it, Screenshot 2025-06-10 at 9.02.58 PM.pngScreenshot 2025-06-10 at 9.03.04 PM.pngScreenshot 2025-06-10 at 9.03.10 PM.pngin these images, example number three shows an example similar to the one I provided, which is f(x)=x^(2/3), at first it says that it doesn't have a vertical tangent, but then it says "In the light of the two preceding examples, we extend the definition of tangent line to allow for vertical tangents" which then provides a new and broader definition that allows the function to have a vertical tangent, I argued with Google Gemini and now I'm convinced that it does actually have vertical tangent, here is the full link for my conversation with Google Gemini


my teacher's argument was that the limit of the derivative must be +infinity or -infinity in both sides at the same time, but in this example the signs of the infinities are different from the both sides, so he said it doesn't have vertical tangent, I would be great full if you told me and re assured me wether this function have a tangent at 0 or not, if yes, is the tangent vertical?

thanks.
 
but then it says "In the light of the two preceding examples, we extend the definition of tangent line to allow for vertical tangents" which then provides a new and broader definition that allows the function to have a vertical tangent
Sounds like an argument about definitions, not about substance: your book extends the definition, but your teacher does not :)
 
Sounds like an argument about definitions, not about substance: your book extends the definition, but your teacher does not :)
I get it. but what I mean is, is the second definition the highest standard for a vertical tangent? is it the highest accepted standard and according to it the function does actually have a vertical tangent unlike what my teacher said? or, there isn't a standard and there are only different definitions for it which lead to different points of view?
 
f(x)=x4/5 f(x)=x^{4/5} has a first derivative f(x)=451x1/5 f'(x)=\dfrac{4}{5} \dfrac{1}{x^{1/5}} which is not defined at x=0, x=0, means that f(x) f(x) is not differentiable at 0. 0.

Your teacher's argument that the limits from both sides have to be equal is correct. And the limits at 0 0 of the function have different signs, which is not allowed.

This is the standpoint of differentiability.

Now, we open an entirely new discussion. The geometry of the function graph {(x,f(x)xR}. \{(x,f(x)\,|\,x\in \mathbb{R}\}. If we draw secants between the points (a,(a)(4,5)) (-a,(-a)^{(4,5)}) and (0,0) (0,0) with a positive value a>0 a>0 and make a a smaller and smaller, then we end up with ever steeper straights resulting in the vertical line x=0. x=0. Same if we do it from the right. Hence, we can geometrically construct such a line. I wouldn't call it a tangent to avoid confusion, but this is only an opinion. The point is, that if we draw secants between the points (a,(a)(4,5)) (-a,(-a)^{(4,5)}) and (+a,(+a)(4,5)) (+a,(+a)^{(4,5)}) and let a a approach zero, then we end up with the line y=0. y=0. That's not what you want to have, two different results depending on how you define the limit. That's why the function isn't differentiable and why I wouldn't say tangent. It is geometry, and only geometry. Geometry doesn't know what limits are, so the transition from secants to tangents is a bit artificial. This is why we choose calculus instead of geometry, and calculus says, f(x) f(x) is not differentiable at x=0. x=0.

My favorite formula is Weierstraß formulation of differentiability. A function is differentuíable at a point x x (in the direction of v v ) if there is a linear function L:RR L\, : \,\mathbb{R}\longrightarrow \mathbb{R} and a remainder function r(v) r(v) that decreases faster to zero than v v such that f(x+v)=f(x)+L(x)v+r(v). f(x+v)=f(x)+ L(x)\cdot v + r(v). This has all it needs to define a derivative or a tangent, namely that linear function L. L. In the case of f(x)=x4/5, f(x)=x^{4/5}, we cannot find such a linear function, hence there is no derivative and no tangent. What you do geometrically is another topic and depends on the choice of your geometric rules as the example shows that delivered x=0 x=0 in one process, and y=0 y=0 in another.
 
Last edited:
Top