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How to prove that the derivative f this function is |x|

  1. Feb 24, 2009 #1
    how to prove that the derivative of this expression


    [tex]
    f(x)=-\frac12x^2,x<0
    [/tex]
    [tex]
    f(x)=\frac12x^2,x>=0
    [/tex]
    is f'(x)=|x|

    i tried
    [tex]
    \lim _{x->0^-}\frac{f(x)-f(0)}{x}=\lim _{x->0^-}\frac{-\frac12x^2-0}{x}=\lim _{x->0^-}-\frac12x=0\\
    [/tex]
    [tex]
    \lim _{x->0^-}\frac{f(x)-f(0)}{x}=\lim _{x->0^-}\frac{+\frac12x^2-0}{x}=\lim _{x->0^+}+\frac12x=0
    [/tex]
    but i get values
    it doesnt show that f'(x)=|x|
    ??
     
    Last edited: Feb 24, 2009
  2. jcsd
  3. Feb 24, 2009 #2

    CompuChip

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    You calculated f'(0) and found that it is 0. So this doesn't contradict f'(x) = |x|. Now calculate the derivative for all other x as well :tongue:
     
  4. Feb 24, 2009 #3
    [tex]
    \lim _{h->0^-}\frac{f(x+h)-f(x)}{h}=\lim _{h->0^-}\frac{-\frac12(x+h)^2+\frac12x^2}{h}=\lim _{h->0^-}\frac{-\frac12(x^2+2xh+h^2)+\frac12x^2}{h}=\lim _{h->0^-}\frac{-\frac12(2x+h)}{1}=-x\\
    [/tex]
    [tex]
    \lim _{h->0^+}\frac{f(x+h)-f(x)}{h}=\lim _{h->0^+}\frac{\frac12(x+h)^2-\frac12x^2}{h}=\lim _{h->0^+}\frac{\frac12(2xh+h^2)}{h}=\lim _{h->0^+}\frac{+\frac12(2x+h)}{1}=x\\
    [/tex]

    what to write in order to finish this prove??
     
  5. Feb 24, 2009 #4

    CompuChip

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    Almost... you didn't mean to write [itex]h \to 0^\pm[/itex] under the limits, because both limits are two-sided. So, the first one is just
    [tex]
    \lim _{h\to0}\frac{f(x+h)-f(x)}{h}=\lim _{h\to0}\frac{-\frac12(x+h)^2+\frac12x^2}{h}=\lim _{h\to0}\frac{-\frac12(x^2+2xh+h^2)+\frac12x^2}{h}=\lim _{h\to0}\frac{-\frac12(2x+h)}{1}=-x\\
    [/tex]
    You probably meant to say, that the top line is for x < 0 and the bottom line is for x > 0 (because on the top line you are using the part of the definition for x < 0 and on the bottom line you are using the definition of f for x >= 0).

    So now you have shown that f' exists everywhere and is equal to
    [tex]f'(x) = \begin{cases} -x & \text{ if } x < 0 \\ 0 & \text{ if } x = 0 \\ x & \text{ if } x > 0 \end{cases}[/tex]

    Can you conclude now that f'(x) = |x| ?
     
  6. Feb 24, 2009 #5
    can I ?
     
  7. Feb 24, 2009 #6

    Tom Mattson

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    Staff Emeritus
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    You should know. Compare his expression for [itex]f'(x)[/itex] with the definition of [itex]|x|[/itex].
     
  8. Feb 24, 2009 #7

    Redbelly98

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    Look at robueler's post concerning y(x) = |x|.
     
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