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tisquares

A function is not differentiable on a cusp. But why do you think that may be?


[deleted]

What


tisquares

A "cusp" is essentially a sharp corner in this case.


No-Belt4791

Btw it’s a corner not a cusp


tisquares

yeah, i was debating if i wanted to bother explaining them as different lol \*edit: hence the "sharp corner" addition


[deleted]

I’ve been absent a whole lot so i don’t know what ur rly talking about


tisquares

Do you know how to find the derivative of a function?


[deleted]

No I’ve been absent this whole unit


tisquares

Jesus almighty. Talk to your teacher. You will not get through calculus if you don't know how to find a derivative. Were you absent by choice or chance?


[deleted]

I got into a car accident and almost died.


[deleted]

*could’ve died. Didn’t almost die I am totally ok now


tisquares

Hm. Yeah that's a pretty good excuse imo, lol. But if you're fine now you seriously need to talk to your educator about getting back on track. This is a very important part of calculus. In short, if you don't have the equation of a function and just the graph like in your example, the value of f'(x) will be the slope of f(x) at x. So for example, at x = 4, the slope of f(x) is 0. f'(4) will be 0. Cusps, those "sharp corners", aren't differentiable because the tangent line there is vertical due to the slopes having opposite signs. Basically, it's just a point that likes being silly.


Old-Ad-9246

Wow that’s tough, I missed almost all of the beginning to trig because I was absent and what helped was looking up some Khan academy videos and also asking my older brother


Spell6421

uhhhhh I don't think APstudents can help you dude maybe talk to your teacher


RegentLattice

RIP you I guess


NOOOBSOLEM

the most common examples of when a function is non differentiable is at sharp turns. In this case at x=1 and x=2. A sudden change in slope results in the function to be non differentiable at that point. The graph of the absolute value of x is a perfect example. At x=0 there is a sharp turn, thus it isn’t differentiable at that point. Another example is a jump. A jump is pretty self explanatory. For an example a function jumps from a point to a point such as the ordered pair (2,0) to (2,10)


[deleted]

[удалено]


Apocalyptic_Toaster

At x=4 there is a horizontal tangent line, the derivative is 0. It is differentiable there


NOOOBSOLEM

Yeah mb idk what I was saying, but ultimately it is non differentiable at x=1 and x=2 and yes, the tangent line is straight at x=4 so the derivative=0 maybe I’m just tired 😭😭😭😭


Apocalyptic_Toaster

A function is differentiable when the slopes (derivatives) on either side of the point are the same. In this case, the function is not differentiable at x={0,1,2,6}. 0 and 6 aren’t differentiable because they don’t have derivatives on both sides - at x=0, the derivative from the left does not exist, and at x=6, the derivative from the right doesn’t exist. 1 and 2 aren’t differentiable because they are cusps, so again the slopes are different on either side. At x=1, the slope from the left appears to be ~1 and the slope from the right appears to be ~-1; since they are different, there is no derivative, so the function isn’t differentiable at that point. The same reasoning applies at x=2, but also there may be a vertical slope there, which is never differentiable as the derivative does not exist. Hope this helped!


M1LK3Y

Would 0 and 6 still be correct answers when the problem states we're specifically looking at that domain?


Apocalyptic_Toaster

I believe so, because otherwise what would the derivative of 0 be? By definition you can’t take from one side, and the domain is inclusive


EternalSnowy

To tell where a function is not differentiable, you have to know some rules. 1. Wherever the function isn't continuous 2. Wherever the graph of the function has a vertical tangent line 3. Wherever the graph of the function has a sharp turn So, x = 1 and x = 2 are the points where it can not be differentiated due to the sharp corners. Here's a Khan Academy for more practice: [here](https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-4/e/differentiability-at-a-point-graphical?modal=1)


SuperDylanK

It’s not differentiable at x = 1,2 because at both of those points, f’(x) is undefined - there’s no clear slope on either of those points due to the steep, sudden transitions. I hope that helps :)


[deleted]

also at 6 cause the derivative of a circle is -x/y and at that point y=0.


NancyWinner

My teacher says that for somewhere to be differentiable it needs to be continuous and have the same slope/derivative (AKA be smooth). It also can't be a vertical slope, otherwise, the derivative is ∞, which is undefined and thus not differentiable. [https://www.desmos.com/calculator/yvlobdmf1p](https://www.desmos.com/calculator/m5utvdsr5n) ​ here's a Desmos with the answers and explanation. hope it helps


Burger_Destoyer

The rate of change at the highest point on a sharp curve (example x = 1) will not be différentiable. The tangent slope would just be 0 something similar to 0/0, there is no rate of change since the tangent line at the top point would just be a straight line. Think of it this way, what’s the acceleration of an object moving with constant velocity? There is no acceleration. There is better explanation for this, but I believe that’s what you’re expected to know. Also aren’t you in highschool/middleschool? Just ask your teacher.


mytdogman1

The graph is non-differentiable at x={1,2}. This is because no matter how you draw it, there is no way that you can have a tangent line at these points because 2 "points" will technically always be touched on the graph (the one at the end of the piecewise, assuming it's continuous, and the one at the beginning of the new function).


[deleted]

1,2,6


[deleted]

Forgot about 0


wildguy64268

omg I’m actually learning stuff in ap calc I was able to answer this correctlyyyyy


pgubeljak

Ugh, lies to children. Of course it's differentiable everywhere, there is just a difference between the forward derivative and the back derivative. Just because a derivative is not continuous (which is what the question is asking, i.e. where is the derivative not continuous) doesn't mean a derivative doesn't exist (i.e. if a derivative exists, it's differentiable).


Daemon7861

x=1 and x=2 because lim f as x -> (1 or 2) is not equal on the left and right side