Derivatives - differentiabilty, continuity, unbounded

In summary: M=|-x^(a-2)cos(1/x)|>M=x^(a-2)|cos(1/x)|>M Now, since x0 is in [0,1], we know x0 is not zero, so x0>0. So x^(a-2)>0. So we need |cos(1/x)|>M/x^(a-2).But cos(1/x) oscillates between -1 and 1, so I don't think there is an x0 in [0,1] such
  • #1
kathrynag
598
0
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:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know xsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 if 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.
 
Physics news on Phys.org
  • #2
kathrynag said:
Let g(x) =xasin(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:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)
I know xsin(1/x) does not exist, but for all other x, g'(0) does exist. Then g'(0) exists and equals 0 if 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.

Hi Kathryn.

Part (a) is rather troublesome.

g(x) is differentiable on R, (except for x = 0, for some values of a). Definitely, for values of a for which g(x) is unbounded on [0, 1], g'(0) does not exist, since in those cases, g(x) is not continuous at x=0.

Your differentiation, g'(x) = ax(a-1)sin(1/x)-x(a-2)cos(1/x) is correct. Of course, this is undefined for x=0.
Another form you can put g'(x) into is:
g'(x) = x(a-2)(a·x·sin(1/x) - cos(1/x))

What you did to check on g'(0), won't work.
Now for x=0, we have:
g'(0)=limg(x)-g(0)/x-0=limg(x)/x=limx^(a-1)sin(1/x)

To determine if g'(0) exists, consider the limit.

limx→0 g'(x).

If this limit exists, you can argue that g is differentiable at x=0.

Graph g(x) and g'(x) for various values of a - I would suggest integer & a few half integer values from a = -1 to a = 4.

In particular look at WolframAlpha.

g(x), a=0 and g'(x), a=0

g(x), a=2 and g'(x), a=2

One sequence to look at is yn = g'(xn), where xn = 1/(π n), especially for a = 2.
 
  • #3
Ok, but I'm still confused on what they mean by unbounded.
 
  • #4
kathrynag said:
...
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.
kathrynag said:
OK, but I'm still confused on what they mean by unbounded.
Unbounded: You can say this in any of several ways.

(xn) is unbounded means: For any M > 0, there exists an n ϵ N, such that |xn| > M .

f(x) is unbounded on [0, 1] means that for any M > 0, there exists an x0 ϵ [0,1] such that |f(x0)| > M .



Did you look at g'(1/(n π))?

g'(1/(n π)) = (1/(n π))(a-2){a·x·sin(1/[1/(n π)]) - cos(1/[1/(n π)])}

= (1/(n π))(a-2){a·(1/(n π))·sin(n π) - cos(n π)}

= - (-1)n(1/(n π))(a-2),     because, sin(n π) = 0, and cos(n π) = (-1)n = ±1 .
If the absolute value of the above sequence diverges for any a, then the sequence and g'(x) are unbounded for that value of a.

If the absolute value of the above sequence converges to any L ≠ 0 for any a, then the sequence and g'(x) are bounded (but g'(x) is not continuous at x=0) for that value of a.

If the absolute value of the above sequence converges to 0 for any a, then the sequence and g'(x) are bounded (and g'(x) is continuous at x=0), for that value of a.
 
  • #5
BTW:

Are you sure that part (a) isn't something like:

g is [STRIKE]differentiable[/STRIKE] continuous on R but such that g' is unbounded on [0,1].

That would make [STRIKE]more[/STRIKE] sense.
 
  • #6
SammyS said:
BTW:

Are you sure that part (a) isn't something like:

g is [STRIKE]differentiable[/STRIKE] continuous on R but such that g' is unbounded on [0,1].

That would make [STRIKE]more[/STRIKE] sense.

No, I'm sure
 
  • #7
So 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?
 
  • #8
kathrynag said:
So 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?
Sorry Kathryn, I've been away for a day.

The answer to your final question in the quoted post is "No." If β>0, then xβ → 0 as x → 0.

Please try g'(1/(n·π)).
 

1. What is the definition of differentiability in calculus?

In calculus, differentiability is a measure of how smoothly a function changes over a given interval. A function is considered differentiable if it has a well-defined derivative at each point within that interval.

2. How is continuity related to differentiability?

Continuity and differentiability are closely related concepts in calculus. A function is said to be continuous if it is defined at every point within a given interval and has no abrupt changes or breaks. A differentiable function is also continuous, but the inverse is not always true. A function can be continuous without being differentiable.

3. Can a function be differentiable at a point but not continuous?

No, a function cannot be differentiable at a point if it is not continuous at that point. Differentiability requires that the function is both defined and continuous at the point in question.

4. What does it mean for a function to be unbounded?

A function is considered unbounded if its values can increase or decrease without limit. In other words, there is no maximum or minimum value that the function can reach within a given interval. This can occur when the function has a vertical asymptote or when it approaches infinity at one or both ends of the interval.

5. How do you determine if a function is differentiable at a specific point?

To determine if a function is differentiable at a specific point, you can use the limit definition of the derivative. If the limit of the difference quotient exists as x approaches the point in question, then the function is differentiable at that point. Additionally, the function must also be continuous at that point in order for it to be differentiable.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
239
  • Calculus and Beyond Homework Help
Replies
2
Views
825
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
436
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
686
  • Calculus and Beyond Homework Help
Replies
2
Views
972
  • Calculus and Beyond Homework Help
Replies
6
Views
342
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
Back
Top