1. ## Differentiable at x=0?

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
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.

2. ## Re: Differentiable at x=0?

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

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

For x < 0, |x| = -x and $f(x) = 1$ and $f(x) = 0$

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!

3. ## Re: Differentiable at x=0?

Originally Posted by tkhunny
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 $(x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1$

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

For x < 0, |x| = -x and $f(x) = 1$ and $f(x) = 0$

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!

4. ## Re: Differentiable at x=0?

How about the derivative from both sides? Exist? Same?

5. ## Re: Differentiable at x=0?

Originally Posted by tkhunny
How about the derivative from both sides? Exist? Same?
Yes, they both would be 0.

6. ## Re: Differentiable at x=0?

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

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

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?

7. ## Re: Differentiable at x=0?

Originally Posted by tkhunny
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

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

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?

8. ## Re: Differentiable at x=0?

It's a whole sentence. The limit is shown.

9. ## Re: Differentiable at x=0?

Originally Posted by LEG7930
Originally Posted by tkhunny
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 $(x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1$

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

For x < 0, |x| = -x and $f(x) = 1$ and $f(x) = 0$

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, $f^\prime(a^+)$ and $f^\prime(a^-)$ ?

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

These left- and right-hand derivatives are the limits of $f^\prime(x)$ 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 $x\neq0$. 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.

10. ## Re: Differentiable at 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?

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