Page 1 of 2 12 LastLast
Results 1 to 10 of 15

Thread: Differentiable at x=0?

  1. #1
    New Member
    Join Date
    Aug 2010
    Posts
    17

    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.

    Thanks for any help you can offer.

  2. #2
    Elite Member
    Join Date
    Apr 2005
    Location
    PA, USA
    Posts
    8,222

    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 [tex](x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1[/tex]

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

    For x < 0, |x| = -x and [tex]f(x) = 1[/tex] and [tex]f`(x) = 0[/tex]

    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. #3
    New Member
    Join Date
    Aug 2010
    Posts
    17

    Re: Differentiable at x=0?

    Quote 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 [tex](x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1[/tex]

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

    For x < 0, |x| = -x and [tex]f(x) = 1[/tex] and [tex]f`(x) = 0[/tex]

    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. #4
    Elite Member
    Join Date
    Apr 2005
    Location
    PA, USA
    Posts
    8,222

    Re: Differentiable at x=0?

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

  5. #5
    New Member
    Join Date
    Aug 2010
    Posts
    17

    Re: Differentiable at x=0?

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

  6. #6
    Elite Member
    Join Date
    Apr 2005
    Location
    PA, USA
    Posts
    8,222

    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

    [tex]f`(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

    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. #7
    New Member
    Join Date
    Aug 2010
    Posts
    17

    Re: Differentiable at x=0?

    Quote 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

    [tex]f`(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}[/tex]

    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. #8
    Elite Member
    Join Date
    Apr 2005
    Location
    PA, USA
    Posts
    8,222

    Re: Differentiable at x=0?

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

  9. #9
    Full Member
    Join Date
    Mar 2009
    Posts
    259

    Re: Differentiable at x=0?

    Quote Originally Posted by LEG7930
    Quote 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 [tex](x+|x|)^{2} + 1 = x^{2} + 2\cdot x\cdot |x| + |x|^{2} + 1 = 2\cdot x^2 + 2\cdot x \cdot |x| + 1[/tex]

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

    For x < 0, |x| = -x and [tex]f(x) = 1[/tex] and [tex]f`(x) = 0[/tex]

    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, [tex]f^\prime(a^+)[/tex] and [tex]f^\prime(a^-)[/tex] ?

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

    These left- and right-hand derivatives are the limits of [tex]f^\prime(x)[/tex] 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 [tex]x\neq0[/tex]. 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.
    -------
    Function Integrator for iPhone - definite integrals in your pocket - http://bit.ly/lpzjZ6

  10. #10
    New Member
    Join Date
    Aug 2010
    Posts
    17

    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?

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •