# Homework Help: Differentiability + Continuity?

1. Feb 17, 2010

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??

for what values of a is f differentiable? is it all a in the real numbers?Dif

2. Feb 17, 2010

### rsa58

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.

3. Feb 17, 2010

### rsa58

oh wait, a >1

4. Feb 17, 2010

so when x=0, a>1 ?

then is a>2 when x is not 0 correct?

5. Feb 17, 2010

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???

6. Feb 17, 2010

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

7. Feb 17, 2010

### snipez90

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,

$$f(x) = x^{a+1} \sin\left(\frac{1}{x}\right),$$

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

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

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

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
8. Feb 18, 2010