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Natural Logs+Calc related question

  1. Mar 6, 2008 #1
    Hi,

    Can someone please help me out and guide me?

    1. The problem statement, all variables and given/known data

    [​IMG]


    2. Relevant equations
    Shown above


    3. The attempt at a solution

    a) f(x) = -2 +ln(x)^2
    f'(x)= 2/x
    Ans: All real numbers except for 0

    b) -2 +ln (x)^2 =0
    ln(x)^2= 2
    e^2=x^2
    Ans: x = +-e

    c) I have no idea how to do this. Do I have to differentiate?


    -Jen
     
  2. jcsd
  3. Mar 6, 2008 #2
    How would you find the "slope of a line that is tangent to the graph"? Once you have the slope, what point would lie on the tangent line if it is tangent to the graph? Once you have the slope, and a point, how do you get the equation of the line?
     
  4. Mar 6, 2008 #3
    OK, so I differentiated the equation and got 2/x
    Then I plugged 1 into x, so our slope is 2 while the y value is --> -2 + 0 = -2

    and the equation i got is --> y+2=2(x-1)

    is that right? Also, are my first two parts right?

    -Jen
     
  5. Mar 6, 2008 #4

    dynamicsolo

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    Homework Helper

    Looks like you're all set on this problem!
     
  6. Mar 6, 2008 #5
    Thanks Mathdope and dynamicsolo!
     
  7. Mar 7, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This part is wrong. If the problem were ln(x2)= 2 then taking the exponential of both sides would give you x2= e2. However, the problem is [ln(x)]2= 2. You need to take the square root of both sides first, then the exponential: ln(x)= [itex]\pm\sqrt{2}[/itex] so x= [itex]e^\sqrt{2}[/itex] or x= [itex]e^{-\sqrt{2}}[/itex].
     
  8. Mar 7, 2008 #7
    Halls, the OP seemed to be the problem you mention but the original image that she uploaded was the other one, so I think it's ok. The poster just needs a bit more care in where parentheses go.
     
  9. Mar 7, 2008 #8

    dynamicsolo

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    Jeez, I didn't even look that closely at what was typed. I was looking at the attachment, so I based my assessment of OP's answers on that version of the problem. (I should be watching what students are typing, but since their answers were correct for the correct statement of the equation, I thought it reasonable that they knew what they were doing...)
     
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