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Math Help - derivative of a function

  1. #1
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    derivative of a function

    Hi, can someone cheak my solution to these problems:

    derivative of a function-untitled.jpg

    For 1 a) i got they don't exist since for a>0- 3a^2
    a<0- -3a^2

    b) again same they don't exist since for a>0- 2a-1
    a<0- -2a+1

    Thsnks

    Ps i put it in this section as i am learing this in an analysis course.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by SubZero View Post
    Hi, can someone cheak my solution to these problems:

    Click image for larger version. 

Name:	Untitled.jpg 
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    For 1 a) i got they don't exist since for a>0- 3a^2
    a<0- -3a^2

    b) again same they don't exist since for a>0- 2a-1
    a<0- -2a+1

    Thsnks

    Ps i put it in this section as i am learing this in an analysis course.
    \lim_{x\to0}\frac{x^2\left| x\right|-0}{x}=\lim_{x\to0}x|x|=0
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    \lim_{x\to0}\frac{x^2\left| x\right|-0}{x}=\lim_{x\to0}x|x|=0
    Hi Drexel28, I thought

    |x| is not differentiable at 0 since |X|/ X =

    1 if X >0
    -1 if X<0
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  4. #4
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    I suggest that you graph both of these.
    I think that you will be surprise by one of them.
    Of course, both have derivatives everywhere except possibly at one point.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by SubZero View Post
    Hi Drexel28, I thought

    |x| is not differentiable at 0 since |X|/ X =

    1 if X >0
    -1 if X<0
    We have that f(x)=x^2|x|, right? So, to claim that f'(0) exists is to claim that \lim_{x\to0}\frac{f(x)-f(0)}{x-0.}=\lim_{x\to0}\frac{x^2|x|}{x}=\lim_{x\to0}x|x| exists. But, -x^2\leqslant x|x|\leqslant x^2 where it follows from the squeeze theorem that \lim_{x\to0}x|x|=0=f'(0).
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