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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

The problem states: if g(x)=x^{2/3}, show that g'(0) does not exist.

So my answer is:

Now lets find left and right limits separately:

From this we can see that left and right limits at point 0 are equal by an absolute value, but have opposite signs; therefore, the

does not exists.But according to the rules of differentiation

or in this case:My problem is: how

correlate with ?They are both supposed to be equal to the slope of tangent line, aren't they? What am I missing?

*Last edited by White_Owl (2012-06-24 14:05:39)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

I don't know if substituting the value of directly into the limit is the right thingto do, but either way, your calculating of the limit is not correct in two ways. First, you reduced it incorrectly to h^(1/3) instead to h^(-1/3). Second you claimed that the left and right limits of what you got aren't the same because they have the same absolute value, but different signs, whereas both limits have the same signed value 0 (which doesn't really have a sign, except when used in limits when denoting left and right limits).

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi White_Owl

It is simpler than that. Take a look at the graph. (i) Values left of x = 0 don't exist (ii) the gradient tends to infinity as x approaches 0 from the right.

And having the two limits (left of point and right of point) is not a test either. Consider the graph

y = -3 x < 0

y = + 3 x > 0

Approaching 0 from either side, the gradients are the same. But the function is undefined and discontinuous at x = 0.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Hi Bob

That is not always the case. Look at the graph of f(x)=x^(3/2). The function is defined for x>0 only, but f'(0) exists and is equal to 0.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

hi Stefy,

It depends on your definition of diffentiation. Some versions require the left limit to equal the right limit and call your example 'semi-diffentiable'. That's why I included the infinite gradient test as well.

http://en.wikipedia.org/wiki/Semi-differentiability

(Actually I agree with you .... but I have met the alternative definition.)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,885

Still, the f'(0) exists at that point.

I am not sure how that now applies on y=x^(2/3) .

But here is another example: x^(5/3). It is differentiable around 0 and the derivative at 0 is 0.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**White_Owl****Member**- Registered: 2010-03-03
- Posts: 99

Ok.

Question #1.

According to Stewart's textbook, Calculus Early Transcendentals 7th edition, we can substitute x for an exact value if we are looking for the slope of the tangent at that point.

Chapter 2.7, Pages 144-145.

And Example #2 does exactly that.

Question #2.

Yes, I did a mistake in the final step of calculating the limits. Sorry about that. It should be:

But still, I have vs .

Where did 2/3 disappeared?

Question #3.

Actually, I do not see why did you say that domain of the function is just positive numbers? I believe the domain is full R.

After all, we can do :

Not all graphing calculators are smart enough for such function. I am playing with QtiPlot and for "x^(2/3)" it draw the similar image as

Which still has a strange horizontal part near the zero, which does not disappear when I zoom in.

*Last edited by White_Owl (2012-06-26 06:47:05)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

Which still has a strange horizontal part near the zero, which does not disappear when I zoom in.

Did you mean 'vertical' ?

That alone would be the proof you want because the limit wouldn't exist at x = 0.

Arhh! seen your graph. That horizontal bit is fake. Most plotters work by evaluating as x is gradually incremented and then joining the resulting points. So what you are seeing is the effect of joining two points across the gap where x = 0. Think about what the limit actually does and you'll see it tends to infinity (-infinity) as x tends to zero.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Jay Bhavsar****Guest**

You are supposed to put (x+h)^(2/3) not (0+h)^(2/3) when finding the derivative using the definitions (f(x+h)-f(x))/h.

Because the limit of h is approaching 0, not x.

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