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Determine when f is differentiable

  1. Dec 9, 2005 #1
    How do you determine when f is differentiably from a real analysis standpoint (no graphs and calculus)? Would I simply look for a point of discontinuity? We have 4 problems on our homework assignment involving this issue and I don't see one example in my notes or the book adressing it. Here is one of our problems incase your wondering:

    Determine the values of x for which f(x) is differentiable and find it's derivative:
    f(x) = x|x|

    Thanks
     
  2. jcsd
  3. Dec 9, 2005 #2

    JasonRox

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    Start by looking at f(x) = |x|, and using the definition of the derivative as x approaches 0.

    Are the limits the same?
     
  4. Dec 9, 2005 #3
    I don't understand. Are which two limits the same? The definition of the derivative on |x| and x|x|?
     
  5. Dec 9, 2005 #4

    D H

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    You misunderstood JasonRox' post. I'll expand on it. First, forget about [itex]x{\lvert x\rvert}[/itex] for a moment. Just look at [itex]{\lvert x\rvert}[/itex]. When you are looking at that function, examine its behavior near zero. In particular, look at
    [tex]\lim_{\Delta x\to 0}\frac{{\lvert x + \Delta x\rvert} - {\lvert x\rvert}}{\Delta x}[/tex]
    at [itex]x=0[/itex].

    By two limits, JasonRox meant the limit as [itex]\Delta x[/itex] approaches zero from below (i.e., make [itex]\Delta x[/itex] some small negative number that approaches zero) versus above (same thing, but now make [itex]\Delta x[/itex] positive).
     
    Last edited: Dec 9, 2005
  6. Dec 9, 2005 #5

    JasonRox

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    Thanks, for clearing it up.

    Yeah, once you see the idea behind f(x) = |x|, you'll totally understand it for f(x) = x|x|.

    Draw the graph as well, so you see what's happening.
     
  7. Dec 9, 2005 #6

    matt grime

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    of course if you remember |x|=x for x=>0 and -x otherwise then it is clear that the functions is differentialbe away from the origin.
     
  8. Dec 9, 2005 #7

    HallsofIvy

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    First of all, NO, it is NOT the case that "if a function is continuous at x= a, it is differentiable at x= a"! For example, f(x)= |x|. That is continuous at x= 0 but not differentiable at x= 0.

    f(x)= x|x|= x2 if x> 0 which is obviously differentiable for x> 0.
    f(x)= x|x|= -x2 if x< 0 which is obviously differentiable for x< 0.

    Now what about
    [tex]lim_{h\rightarrow 0^+}\frac{f(h)- f(0)}{h}[/tex]
    and
    [tex]lim_{h\rightarrow 0^-}\frac{f(h)- f(0)}{h}[/tex]
    ?
    Take the limit
     
  9. Dec 9, 2005 #8

    HallsofIvy

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    First of all, NO, it is NOT the case that "if a function is continuous at x= a, it is differentiable at x= a"! For example, f(x)= |x|. That is continuous at x= 0 but not differentiable at x= 0.

    f(x)= x|x|= x2 if x> 0 which is obviously differentiable for x> 0.
    f(x)= x|x|= -x2 if x< 0 which is obviously differentiable for x< 0.

    Now what about
    [tex]lim_{h\rightarrow 0^+}\frac{f(h)- f(0)}{h}[/tex]
    and
    [tex]lim_{h\rightarrow 0^-}\frac{f(h)- f(0)}{h}[/tex]
    ?
    Take the limit
     
  10. Dec 9, 2005 #9

    JasonRox

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    Yeah, totally right.

    It is true that if it is differentiable at x=a, then it is continuous at x=a.

    The converse is not true, just like HallsofIvy showed.
     
  11. Dec 10, 2005 #10
    Ok, I did what you guys said and took the right and left hand derivatives for this function.
    I got the right hand derivative to be infiinity, while the left is -infinity. So, it's not differentiable at x = 0. I understand this, and I used this technique on the 4 other problems. I have one question, however.

    For the function f(x)=|x^2-1|, I found it to be differentiable at x = -1, but not at x = 1. Intuitively, this is wrong, but I can't find my mistake. The right and left hand derivatives I got to equal 0. Is this correct?
     
  12. Dec 10, 2005 #11

    matt grime

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  13. Dec 10, 2005 #12
    Thanks for the help guys, I made some pretty dumb mistakes in a few of my posts that I caught tonight after looking things over. I think I have the answers now. when f(x) = |x^2-1|, it's not differentiable at x = 1, -1. when f(x) = x|x|, it's differentiable at 0. Is this right? If needed, ill post work so you guys just don't think your giving me an answer.
     
  14. Dec 10, 2005 #13

    JasonRox

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    Just plot that the graph.

    That's the easiest way to know.
     
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