Differentiable at x=0?

LEG7930

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So the question is: Decide if the function is differentiable at x=0.

f(x)=((x+abs(x))^2)+1

My first instinct is to find the derivative of f(x) and then plug in x=0. My only problem is HOW do you find the derivative of an absolute value?

Since I didn't know how to find the derivative (hopefully someone can tell me the rule for absolute values) I did however use the derivative calculator on this website and found it to be
ht5soj.gif

But then if you plug in zero you get a 0/0 situation at x/abs(x). So would that mean it is not differentiable at x=0? Because my textbook says that it is.

Thanks for any help you can offer.
 
This is scary: "if you plug in zero you get a 0/0 situation at x/abs(x)." Why would you "plug in" ANYTHING without first determining if it exists? Never do that. You do not get a "0/0 situation", because you never would think of making such a substitution with a value your not sure is in the Domain. Clearly you are not sure. Don't do it!

You just have to wade through it. There is nothing magic. You don't have to expand first, but I decided to do it.

Expand \(\displaystyle (x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1\)

For x > 0, |x| = x and \(\displaystyle f(x) = 4\cdot x^{2} + 1\) and \(\displaystyle f`(x) = 8\cdot x\)

For x < 0, |x| = -x and \(\displaystyle f(x) = 1\) and \(\displaystyle f`(x) = 0\)

You decide what to do for x = 0.
1) Does the limit exist from both sides?
2) If so, is it the same from both sides?

There's a reason why we make and teach definitions. Use them!
 
tkhunny said:
This is scary: "if you plug in zero you get a 0/0 situation at x/abs(x)." Why would you "plug in" ANYTHING without first determining if it exists? Never do that. You do not get a "0/0 situation", because you never would think of making such a substitution with a value your not sure is in the Domain. Clearly you are not sure. Don't do it!

You just have to wade through it. There is nothing magic. You don't have to expand first, but I decided to do it.

Expand \(\displaystyle (x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1\)

For x > 0, |x| = x and \(\displaystyle f(x) = 4\cdot x^{2} + 1\) and \(\displaystyle f`(x) = 8\cdot x\)

For x < 0, |x| = -x and \(\displaystyle f(x) = 1\) and \(\displaystyle f`(x) = 0\)

You decide what to do for x = 0.
1) Does the limit exist from both sides?
2) If so, is it the same from both sides?

There's a reason why we make and teach definitions. Use them!

OK, so I found the limit as x approaches 0 to be 1, from both sides. Am I correct to use (4x^2)+1 when looking at it from the right and 1 for looking at it from the left? Also, is it the limit that tells whether it is differentiable?

My textbook seems to be confusing me since it doesn't give any sample problems similar to this one. In the book though it says
"A function is differentiable at a point if it has a derivative there." That is why I was plugging in 0 into f'(x).

When you say, " Why would you "plug in" ANYTHING without first determining if it exists?" What are you determining to exist? The limit? If so why?

Also when you say "There's a reason why we make and teach definitions." Which definitions are you talking about so I know to use them.

Sorry if I seem to be completely confused, it's probably because I am at this point. :(

Thanks so much for your help!
 
How about the derivative from both sides? Exist? Same?
 
Okay, let's read that definition again.

Let y = f(x) be a function.

The derivative of f is the function whose value at x is the limit

\(\displaystyle f`(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

provided this limit exists.

If this limit exists for each x in an open interval I, then we say that f is differentiable on I.

What do you think? Have we met the requirements?
 
tkhunny said:
Okay, let's read that definition again.

Let y = f(x) be a function.

The derivative of f is the function whose value at x is the limit

\(\displaystyle f`(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\)

provided this limit exists.

If this limit exists for each x in an open interval I, then we say that f is differentiable on I.

What do you think? Have we met the requirements?

"The derivative of f is the function whose value at x is the limit" Is the limit of what? f(x)?

Because if I am understanding correctly, the limit of f'(x) as x goes to 0 is 0. Does that mean it is differentiable?
 
LEG7930 said:
tkhunny said:
This is scary: "if you plug in zero you get a 0/0 situation at x/abs(x)." Why would you "plug in" ANYTHING without first determining if it exists? Never do that. You do not get a "0/0 situation", because you never would think of making such a substitution with a value your not sure is in the Domain. Clearly you are not sure. Don't do it!

You just have to wade through it. There is nothing magic. You don't have to expand first, but I decided to do it.

Expand \(\displaystyle (x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1\)

For x > 0, |x| = x and \(\displaystyle f(x) = 4\cdot x^{2} + 1\) and \(\displaystyle f`(x) = 8\cdot x\)

For x < 0, |x| = -x and \(\displaystyle f(x) = 1\) and \(\displaystyle f`(x) = 0\)

You decide what to do for x = 0.
1) Does the limit exist from both sides?
2) If so, is it the same from both sides?

There's a reason why we make and teach definitions. Use them!

OK, so I found the limit as x approaches 0 to be 1, from both sides. Am I correct to use (4x^2)+1 when looking at it from the right and 1 for looking at it from the left? Also, is it the limit that tells whether it is differentiable?

My textbook seems to be confusing me since it doesn't give any sample problems similar to this one. In the book though it says
"A function is differentiable at a point if it has a derivative there." That is why I was plugging in 0 into f'(x).

When you say, " Why would you "plug in" ANYTHING without first determining if it exists?" What are you determining to exist? The limit? If so why?

Also when you say "There's a reason why we make and teach definitions." Which definitions are you talking about so I know to use them.

Sorry if I seem to be completely confused, it's probably because I am at this point. :(

Thanks so much for your help!

Have you learned about left- and right-handed derivatives? That is, \(\displaystyle f^\prime(a^+)\) and \(\displaystyle f^\prime(a^-)\) ?

And have you learned that if \(\displaystyle f^\prime(a^+)\) and \(\displaystyle f^\prime(a^-)\) both exist and are equal, then the derivative \(\displaystyle f^\prime(a)\) exists and equals the left- and right-hand derivatives?

These left- and right-hand derivatives are the limits of \(\displaystyle f^\prime(x)\) as x approaches a from above and below.

It is correct to say :

* For x>0, f(x) = ((x+abs(x))^2)+1 = 4x^2+1, and f'(x)=8x.
* For x<0, f(x) = 1, and f'(x)=0.

Then, you prove that the limits of f'(x) as x approaches 0 from above and below are the same.

Your derivative calculator has used the fact that |x|' = x/|x| for \(\displaystyle x\neq0\). This is fine, except it didn't tell you that its answer is wrong if x=0. Since you want the derivate when x=0, you'll jhave to take the limit of the calculator's answer as x->0.
 
I have learned about left and right handed limits if that is what you mean.

Let me see if I am understanding though...

So in general, if you are determining whether a point is differentiable you are finding the limit of the derivative from left and right to that point, and if they are equal then the derivative exists, meaning that point is differentiable?
 
LEG7930 said:
I have learned about left and right handed limits if that is what you mean.

Let me see if I am understanding though...

So in general, if you are determining whether a point is differentiable you are finding the limit of the derivative from left and right to that point, and if they are equal then the derivative exists, meaning that point is differentiable?
I think you may have it, but your vocabulary is going to trip you up sooner or later.
You cannot say that you are finding the limit of the derivative of the point a if the derivative of the function does not exist at a.
Instead, you find whether the limits from left and right of {[f(x + h) - f(x)] / h} as h approaches 0 exist at x = a and are equal. If so that limit IS the derivative of f(x) at x = a. The vocabulary may look awkward, but it is exact and will prevent mistakes
 
OK, so just to make sure I understand how about a simple example.
2mhv990.png

Am I doing that correctly?

Just making sure I am at least following what you are saying. If I am, I will attempt to do the original problem again.
 
You ABSOLUTELY have the concept, but I suspect a purist may correctly object to your notation.

I don't think FORMALLY that you can say f'(5) = lim ..... until you show that the limit exists.

So it might be better to say g(5, h) = {[f(5 + h) - f(5)] / h}, where h does not equal 0 and f(x) = 3x[sup:2ahn3qm5]2[/sup:2ahn3qm5].
Then proceed EXACTLY as you did until you you get g(5, h) = 30 + 3h.
Now lim as h -> 0 of [30 + 3h] = does exist and = 30 so f'(5) does exist and = 30.

I am not a mathematician and not too sharp on formalism, but I suspect your answer might get marked down a bit and mine would be marked down less if at all.

This is not NEARLY SO IMPORTANT as that you do have the CONCEPT.
Check the site again after a while to see if the real mathematicians such as tkhunny, or galactus, or lookagain correct my answer.
 
Great, thanks so much for your help!

Now one more question on this topic. If I have a piecewise function, what function would I use to take the lim as h approaches zero.

f(x)=(x^2)sin(1/x) for x not equal to 0
f(x)=0 for x=0

and again it is asking if it is differentiable at x=0.

So I would start with f'(x)= lim as h->0 [f(x+h)-f(x)]/h
But which part of the piecewise would I be using?
 
LEG7930 said:
Great, thanks so much for your help!

Now one more question on this topic. If I have a piecewise function, what function would I use to take the lim as h approaches zero.

f(x)=(x^2)sin(1/x) for x not equal to 0
f(x)=0 for x=0

and again it is asking if it is differentiable at x=0.

So I would start with f'(x)= lim as h->0 [f(x+h)-f(x)]/h
But which part of the piecewise would I be using?
I am going to warn you that I do not usually comment on the calculus page because I am VERY VERY rusty.

First, with a piecewise function, determining differentiability may well require concern about whether the right and left limits are equal.
Second, this specific problem does not seem to me to raise the right/left issue, but being careful with the notation is very important here.

Let's start with the function g(x, h) = {[f(x + h) - f(x)] / h} where h does not equal 0, f(x) = 0 when x = 0, and f(x) = x^2 * sin(1/x) for x not equal 0.
Notice that we have not at this point made an suppositions about whether limits exist or not and have not said anything about f'(x).
So, g(0,h) = {[f(0 + h) - f(0)] / h} = {[f(h) - 0] / h} = f(h) / h = {[(h^2) * sin(1/h)] / h} = [h * sin(1/h)].
Now, the question is whether the limit of [h * sin(1/h)] as h approaches 0 exists and, if so, what it equals.
Your view?
 
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