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Differentiable function

  1. Jul 4, 2011 #1
    f(x)= x=1/x-1 if x ≤ 0
    x^2-2x +1 if 0 < x < 1
    ln x if x ≥ 1

    Here's what I have so far
    lim f(x)= lim x+1/x-1= -1, LHL= -1
    x →0- x→0-
    lim f(x)= lim x^2-2x+1= (1/2)^2 - [(2)1/2] +1= 1/4, RHL= 1/4
    x→0+ x→0+
    RHL ≠ LHL ∴ f(x) is not differentiable @ 0

    Now I know I need to test x=1 for differentiability, but I am stuck on how to do that
    I can take f ' (x)= 1/x which is continuous for all except 0, but not sure how to test limits
    since there is no LHL with x ≥ 1
     
  2. jcsd
  3. Jul 4, 2011 #2

    SammyS

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    Use of parentheses is needed for your expressions to say what I think you mean for them to say.

    I think you mean:

    f(x) =
    (x+1)/(x-1) , if x ≤ 0
    x2-2x +1 , if 0 < x < 1
    ln x , if x ≥ 1​

    Your limit for x→0+ is not correct. It's not 1/4.
    [tex]\lim_{x\to 0^+}(x^2-2x +1) = 1[/tex]

    What is [itex]\displaystyle \lim_{\,x\to 1^-}(x^2-2x +1)\ \ ?[/itex]

    What is [itex]\displaystyle \lim_{\,x\to 1^+}\ln(x)\ \ ?[/itex]
     
  4. Jul 4, 2011 #3
    Thanks, you are correct about the setup of f(x).
    I still get 1/4 when I plug 1/2 into x^2 - 2x + 1
    1/2^2= 1/4, 2(1/2)=1
    1/4 -1 + 1= 1/4
    The limits set up for 1 helps, for some reason I thought I couldn't use the polynomial, but I finally found it in my notes.
    Thank you
     
  5. Jul 4, 2011 #4
    nvrmind previous post about RHL, I see my mistake.
    So,
    for the second part, I get
    lim (x^2 - 2x +1)= 0
    x→1-

    lim lnx= 0
    x→1+
    Which means f(x0 is differentiable at 1 and final answer is
    f(x) is differentiable for all real numbers x≠0
    Correct?
     
  6. Jul 4, 2011 #5

    SammyS

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    Yes, That is correct !!!
     
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