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Differentiability + Continuity?

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose a>0 is some constant and f:R->R is given by
    f(x) = |x|^a x sin(1/x) if x is not 0
    f(x) = 0 if x=0

    for which values of a is f differentiable at x=0? Use calculus to determine f'(x) for x is not equal to 0. For what values of a is f' a continuous function defined on R? Justify your answer

    2. Relevant equations



    3. The attempt at a solution
    when f is not 0, f'(x) = a|x|^(a-1) x sin (1/x) - a.|x|^(a-2) x cos (1/x)
    so f' is continuous for a>2 where a is a natural number??

    what about when x=0?
    for what values of a is f differentiable? is it all a in the real numbers?Dif
     
  2. jcsd
  3. Feb 17, 2010 #2
    apply the definition of the derivative at x=0. set y=1/x and then take the limit as y goes to infinity. this limit exists of a >=2 otherwise it obviously does not. i think this is true check it.
     
  4. Feb 17, 2010 #3
    oh wait, a >1
     
  5. Feb 17, 2010 #4
    so when x=0, a>1 ?

    then is a>2 when x is not 0 correct?
     
  6. Feb 17, 2010 #5
    when x is not equal to 0, do we have to consider x>0 and x<0?
    will that make a difference to the value of a?
    so is a>1 or a>=2? and a has to be a natural number right? my friends and I have been discussing this and we don't know who is right...

    what about when x=0? is f(x)=0 so a can be anything???
     
  7. Feb 17, 2010 #6
    oh and is it correct to use the algebra of continuous function?
    after differentiating f,
    assuming cos and sin are continuous function, cos + sin is continuous
    we know that x^n where n is a natural number including 0 is continuous
    therefore consider the power to the x
    which is a-2 and a-1
    then a-2 >= o and a-1>=o
    we have a>=2 and a>=1
    then choose maximum and get a>=2
    and a>=2 is the same as a>1 right? if we consider a to be a natural number
     
  8. Feb 17, 2010 #7
    Um what, are you guys guessing values of a.

    First of all, your expression for f'(x) when x is not 0 is incorrect. When x > 0,

    [tex]f(x) = x^{a+1} \sin\left(\frac{1}{x}\right),[/tex]

    and you need to differentiate this expression to determine f'(x) for x > 0.

    For differentiability at 0, apply the definition of f'(0)

    [tex]f'(0) = \lim_{x\rightarrow 0} \frac{f(x) - f(0)}{x} = \lim_{x\rightarrow 0} \frac{|x|^a x\sin\left(\frac{1}{x}\right)}{x} = \lim_{x\rightarrow 0} |x|^a \sin\left(\frac{1}{x}\right).[/tex]

    Try proving rigorously that this last limit is 0 for any a > 0 (the squeeze theorem would be good here). Note that generally, a is never assumed to be a natural number in these kinds of problems.
     
    Last edited: Feb 17, 2010
  9. Feb 18, 2010 #8
    I've forgotten to consider 2 cases when x>0 and x<0
    when x>0, f' = -x^(a-2) cos (1/x) + ax&(a-1) sin (1/x)
    when x<0, f' = x^(a-2) cos (1/x) - ax&(a-1) sin (1/x)

    are these correct? do we apply algebra of continuous function here?
    and that a>1 or a>=2?
    are these two equivalent?
     
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