Empty
  • Empty
Is it possible to give the derivative of |x| at x=0 a value?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
no, the derivative is undefined at x = 0
anonymous
  • anonymous
the derivative as x approaches zero from the right is 1 , the left hand slope is -1
Empty
  • Empty
Yes the algorithm doesn't define a derivative there, but look at this graph: https://www.desmos.com/calculator/cwbnewkrud We can see that in order to turn through that single point the derivative there looks like it should, if it did have a value from the algorithm, take on a value between -1 and 1. Not only that, it's at a local min, which normally means that the derivative should be zero there, which also lies between -1 and 1. Just because it doesn't approach it doesn't mean we can't define the derivative to be 0 there since it seems to fit. It seems more useful than leaving it undefined and it makes sense to do so I think.

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anonymous
  • anonymous
What algorithm?
Empty
  • Empty
This one: \[f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\]
anonymous
  • anonymous
for h>0 : (f(0 + h ) - f(0)) / h = f(h)/h = h / h = 1 , and lim 1 = 1 for h <0 (f(0 + h ) - f(0)) / h = f(h)/h = -h / h = -1, and lim -1 = -1
Empty
  • Empty
Right, I'm not really debating that the algorithm doesn't define anything there, but I did give two good reasons why we should expand the concept to say the derivative should be 0 at x=0.
anonymous
  • anonymous
that limit does not exist at x = 0 , so the derivative is undefined. $$ \large f'(x) = \lim_{h \to 0^{+}} \frac{f(x+h)-f(x)}{h}= \lim_{h \to 0^{-}} \frac{f(x+h)-f(x)}{h} $$
anonymous
  • anonymous
the derivative must equal the right hand limit and the left hand limit if it is to exist
Empty
  • Empty
I don't think you're really paying attention to what I'm saying.
anonymous
  • anonymous
We have an accepted definition of derivative. I don't know why you want to violate it.
freckles
  • freckles
Your picture shows there isn't a unique tangent line at x=0.
geerky42
  • geerky42
^
Empty
  • Empty
Yeah it shows that at that point the slope could possibly be any value in a range from -1 to 1. But the slope is definitely not -3 or 7 there. Imagine a particle flying along that path, it will be the tangent vector. Right when it gets to x=0 it will have to turn through an angle before it can continue flying straight. I'm saying that the regular derivative definition seems like it might be too simple or naive. What if we need something better? It seems to have a range there. My other argument is that this is a local min, and so commonly we find the derivative is 0 at a local min and max, so why not take this geometric stance in addition to or separately to expand it? At the very least, what makes everyone so sure that limit definition is the one and only true and final derivative definition that can never be improved upon? What's stopping us from potentially revising and generalizing it perhaps? I don't have the answer to this question but it feels to me like you all have ruled this possibility out -- why?
freckles
  • freckles
local max/min can occur when f'=0 or when f' dne
freckles
  • freckles
because of the definitions already here we can make a new definition maybe
anonymous
  • anonymous
You can show that f'(0) cannot be defined. Assume f'(0) exists \[ \Large \text{Definition} \\ \large{ f'(x) = \lim_{h \to 0^{+}} \frac{f(x+h)-f(x)}{h}= \lim_{h \to 0^{-}} \frac{f(x+h)-f(x)}{h} \\~\\ f' (0) = \lim_{h \to 0^{+}} \frac{f(0+h)-f(0)}{h} = 1 \\ f' (0) = \lim_{h \to 0^{+}} \frac{f(0+h)-f(0)}{h} = -1 \\ \therefore \\ 1 = -1 } \]
Empty
  • Empty
Yeah, please don't argue this same thing again @jayzdd I understand completely what you're saying, it's just not addressing my issue.
anonymous
  • anonymous
You could make an alternate definition of derivative, but we have a lot of theorems that depend on the limit definition of derivative. So you would have to start from scratch in light of this new derivative. In topology there is an other way to define derivative, so this is not the last word on derivative.
freckles
  • freckles
g star of a function g: \[g^{*}(x)=g'(x) \text{ if } g'(x) \text{ exist } \\ g^{*}(x)=0 \text{ if } g'(x) \text{ doesn't exist } \] so g(x)=|x| g star=x if x>0 and -x if x<0 and 0 if x=0 |dw:1433873776520:dw| now why did we need this?
freckles
  • freckles
and I should say if g'(x) doesn't approach one of the infinities
freckles
  • freckles
g star=1 if x>0 and -1 if x<0 and 0 if x=0 *
freckles
  • freckles
This might sound better: \[\text{ Assume } g \text{ is continuous } \\ g^*(x)= \left\{ \begin{matrix}g'(x) \text{ if } g'(x) \text{ exist } \\ 0 \text{ if } g'_{_{-}}(x)=L \text{ and } g'_{_{+}}(x)=M, L \neq M \end{matrix} \right\} \] where M and L are real numbers and g'_ is left derivative and g'+ is right derivative
anonymous
  • anonymous
I am thinking about looking at f(x) = x^(2/3). If we want to define the derivative at x =0, now you can pick any slope between negative infinity and positive infinity
anonymous
  • anonymous
I don't think its prudent to force a non smooth function to have a derivative at all points. Because when we say that a function has a well defined derivative at all points in its domain, then we expect it to be smooth all points in its domain, not have a kink as in y=|x|. For example y = x^2 has a derivative at all points. There are many theorems that would have to change if you deviated from this.
anonymous
  • anonymous
@Michele_Laino Would you like to comment on this problem?
Empty
  • Empty
Another argument is if we consider the Fourier Series of the sgn function it converges to 0 at x=0. So there is a more "natural" way in which this happens.

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