Re: help with a basic differentiability and derivative function


"David Wilkinson" <david@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:dob12k$m29$1@xxxxxxxxxxxxxxxxxxxxxx
> David L. Wilson wrote:
>> I am replying to my own post to add an additional final paragraph to
>> further clarify.
>>
>> "David L. Wilson" <dwilson314@xxxxxxxxxxxx> wrote in message
>> news:Y6Gdna_kruMIWTXenZ2dnUVZ_sOdnZ2d@xxxxxxxxxxxxxxx
>>
>>>"David Wilkinson" <david@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
>>>news:do92n6$p1r$1@xxxxxxxxxxxxxxxxxxxxxxx
>>>
>>>>David L. Wilson wrote:
>>>>
>>>>>"David Wilkinson" <david@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in
>>>>>message news:do8f40$pjb$1@xxxxxxxxxxxxxxxxxxxxxxx
>>>>>
>>>>>
>>>>>>disanalysis wrote:
>>>>>>
>>>>>>
>>>>>>>very basic problem...can anyone help me with the proof?
>>>>>>>
>>>>>>>f(x) = x^2-3x+8 if x>1
>>>>>>>f(x) = 2x^2+x+3 if x=<1
>>>>>>>
>>>>>>>using epsilon-delta property can you prove that f(x) is
>>>>>>>differentiable
>>>>>>>at all points except x=1
>>>>>>>and that it is not differentiable at 1?
>>>>>>>
>>>>>>>thanx
>>>>>>>
>>>>>>
>>>>>>At x = 1 f(x) = 6 from both expressions.
>>>>>>
>>>>>>for x > 1, df(x)/dx = 2x-3 -> -1 as x -> 1 from the first expression,
>>>>>>and
>>>>>>
>>>>>>For x <= 1, df(x)/dx = 4x+1 = 5 at x=1.
>>>>>>
>>>>>>So the first derivative of f(x) is discontinuous at just above x = 1.
>>>>>>But, at x = 1 it is differentiable from the second expression which
>>>>>>holds up to and including x = 1.
>>>>>
>>>>>
>>>>>Wrong. The derivative does NOT exit at x=1 because the limit
>>>>>definining the derivative at that point does not exist as the limit
>>>>>defining the derivative at x=1 from the left is not equal to the limit
>>>>>defining the derivative at x=1 from the right. Or were you
>>>>>intentionally giving him a wrong answer on this obvious homework
>>>>>problem.
>>>>>
>>>>>Graphically, there is a sharp turn at x=1, hence, the derivative does
>>>>>not exist at the point (1, 6) although the function is continuous
>>>>>there.
>>>>
>>>>Right. Since the second expression holds up to x = 1 the derivative is
>>>>defined by it at x = 1. The first expression only holds for x > 1 so the
>>>>derivative is different for x > 1, not at x = 1.
>>>
>>>NO NO NO
>>>The derivative at a point is defined as a limit (the usual definition in
>>>a calculus text). That limit exists if and only if that limit from the
>>>right and that limit from the left are the same. In this case, that
>>>limit from the left is 5 and that limit from the right is -1 so the limit
>>>that defines the derivative at x=1 does not exist. However, one can say
>>>at x=1, the left derivative is 5 and the right derivative is -1.
>>>
>>>The fact that one of his equations defines the function for x<=1 does not
>>>allow one to use it alone to define the derivative at x=1----Your
>>>sentence "Since the second expression holds up to x = 1 the derivative is
>>>defined by it at x = 1." is terribly wrong.
>>
>>
>> To look at it another way...Change the definition of the function to
>> f(x) = x^2-3x+8 if x=>1 (I moved the equal to here.)
>> f(x) = 2x^2+x+3 if x<1
>> This is the same function. Each x has the same f(x) value. But by your
>> argument the derivative at x=1 is now -1 instead of 5---but it is the
>> same function. Something must be wrong (namely the derivative at x=1
>> actually does not exist).
>>
>>
> No, that is not the same function. You have changed the value of f(x) at x
> = 1 from 2x^2+x+3 to x^2-3x+8

Which are both 6. A function by definition is a rule that assigns a value
in the range to each value in the domain. Two function are equal (the same)
if the rules assign the same values at all points (these do). Or
equivalently if they have the same graph (points (x,y)) which these do.
The piecewise function defined above is the same function as the earlier
piecewise defined function.

> so it is a different function with a different slope at x = 1.

The definition of a function does not care about slope or derivatives.

> As I understand it, if a function g(x) has continuous ordinate and first
> derivative at x

"continuous" not require above as having a first derivative implies
continuity itself.

> then one can define the first derivative as any of
>
> Lim delx -> 0 (g(x+delx)-g(x))/((x+delx)-x) or
>
> Lim delx -> 0 (g(x)-g(x-delx))/(x-(x-delx)) or
>
> Lim delx -> 0 (g(x+delx)-g(x-delx))/((x+delx)-(x-delx))
>
> and they would all give the same answer.

True with the understanding that delx can be positive or negative (whcih you
must allow if you are going to use any one of them as the definition).

> However, in this case the first derivative of f(x) is discontinuous at
> just above x = 1,

"just above" - mathematically there is no such thing. The derivative is
discontinuous at x=1 because it does not exist at x=1 so the limit of the
derivative as x approaches x=1 cannot be (as required for continuity) the
derivative at x=1. It is true though that the derivatives from the right
and from the left do exist at x=1 (and being not equal, the derivative does
not exist at x=1).

You may want to visit one of your local Ph.D.'s in mathematics. You will
find he will tell you the same thing that I am saying--this is standard
stuff.





.

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