Let g(x) =x^asin(1/x) if x is not 0
g(x)=0 if x=0
Find a particular value for a such that
a) g is differentiable on R but such that g' is unbounded on [0,1].
b) g is differentiable on R with g' continuous but not differentiable at zero
c) g is differentiable on R but and g' is differentiable on R, but such that g'' is not continuous at zero.
I've only started on a) but I got a little stuck
Well I know that a function is differentiable if g' exists for all points.
I started by calculating derivatives:
g'(x)=ax^(a-1)sin(1/x)-x^(a-2)cos(1/x) if x is not zero.
Now for x=0, we have:
I know limxsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 provided a>2.
So g is differentiable provided a>2.
Now the part that confuses me is the unboundedness. I looked in this particular chapter of my book and there was no mention of boundedness at all.
So from previous chapters, I had that a sequence (xn) is bounded if there exists a number M>0 such that |xn|<=M for all n in N. I just am really confused on the unboundedness part.
g' unbounded means we want for M>0, there exists x0 in [0,1] such that |g'(x0)|>M
We had g'(x) = ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)
Now we want |ax^(a-1)sin(1/x)-x^(a-2)cos(1/x)|>M
We know limx^(a-1)sin(1/x)=0, so we are only concerned with -x^(a-2)cos(1/x)
So to be unbounded, we want a>2?