Re: help with a basic differentiability and derivative function


"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.


.

Relevant Pages

  • Re: help with a basic differentiability and derivative function
    ... at x = 1 it is differentiable from the second expression which holds up to and including x = 1. ... 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 ... if a function ghas continuous ordinate and first derivative at x then one can define the first derivative as any of ... expression is not useful in this case because of the discontinuity at just above x = 1. ...
    (sci.math.num-analysis)
  • Re: help with a basic differentiability and derivative function
    ... David L. Wilson wrote: ... 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. ... Graphically, there is a sharp turn at x=1, hence, the derivative does not exist at the point although the function is continuous there. ...
    (sci.math.num-analysis)