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Math Help - differentiating absolute value graphs

  1. #1
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    differentiating absolute value graphs

    hi, im a bit confused on the graph of y=abs(lnx)

    when i graph it, my answer only appears in the first quadrant since x>0, and where the negative starts reflecting on the x axis is the c.p so i cant differentiate that spot. now when i check my graph on wolfram alpha, it says that the graph is in both the 2nd and 1st quadrant - Wolfram|Alpha
    could anyone explain why please?
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  2. #2
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    Quote Originally Posted by purebladeknight View Post
    hi, im a bit confused on the graph of y=abs(lnx)

    when i graph it, my answer only appears in the first quadrant since x>0, and where the negative starts reflecting on the x axis is the c.p so i cant differentiate that spot. now when i check my graph on wolfram alpha, it says that the graph is in both the 2nd and 1st quadrant - Wolfram|Alpha
    could anyone explain why please?

    You wrote the function y=\left|\ln x\right| , and this function's defined only for x>0, so I don't know who or what is telling you that that its graph is in both the first and second quadrants: it/he/she is wrong.
    Since by definition |x|:=\left\{\begin{matrix}x&\mbox{ , if }x\ge 0\\-x&\mbox{ , if }x<0\end{matrix}\right. , we get y=|\ln x|=\left\{\begin{matrix}\ln x&\mbox{ , if }\ln x\ge 0\\-\ln x&\mbox{ , if }\ln x<0\end{matrix}\right. =\left\{\begin{matrix}\ln x&\mbox{ , if }x\ge 1\\-\ln x&\mbox{ , if }0\le x<1\end{matrix}\right.

    Thus, the graph of this function is a mirror-reflexion thru the x-axis of the graph of \ln x, for 0\le x<1, and it is exactly the same as the graph of \ln x for  x> 1 .

    Tonio
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