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Very simple derivative problem don't know what I'm doing wrong!

  1. Oct 29, 2011 #1
    Case closed!

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

    I'm reviewing for a test and I'm surprised of this. I've did countless problems harder than this.. for some reason every time I use the constant rule I get it wrong.. I want to understand.. WHY?!?

    Determine the points at which the graph of the function has a horizontal line.

    [tex]g(x)=\frac{8(x-2)}{e^x}[/tex]

    3. The attempt at a solution

    [tex]g(x)=\frac{8(x-2)}{e^x}[/tex]
    By the constant rule:
    [tex]=8 (\frac{x-2}{e^x}[/tex]
    Definition of quotient rule with constant
    [tex]c \frac{\frac{d}{dx} f(x) g(x) - \frac{d}{dx} g(x) f(x)}{[g(x)]^2}[/tex]
    [tex] 8 \frac{1(e^x)-e^x(x-2)}{e^{2x}}[/tex]
    Distribute [tex]e^x[/tex]
    [tex]=8 \frac{e^x - xe^x - 2e^x}{e^{2x}}[/tex]
    Divide by [tex]e^x[/tex]
    [tex]= 8 \frac{-x-2}{e^x}[/tex]
    [tex]= \frac{-8x-16}{e^x}[/tex]
    Horizontal line (0 slope) is at g'(-2)
    [tex] g(2)= \frac{-8(x-2}{e^x}[/tex]
    [tex] g(2) = \frac{-32}{e^-2}[/tex]
    [tex] g(2) = -32e^2[/tex]
    So g(x) has a horizontal tangent at [tex](-2,-32e^2)[/tex]

    But the right answer is supposed to be horizontal line slope at g'(3) and horizontal tangent at [tex](3,\frac{8}{e^3})[/tex]

    What am I doing wrong? Am I using the constant rule the wrong way? It makes me go crazy because I know I have all the basic derivative rules in check. Please help!
     
    Last edited: Oct 29, 2011
  2. jcsd
  3. Oct 29, 2011 #2

    SammyS

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    It's OK to here.
    [itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x+2\,.[/itex]
     
  4. Oct 29, 2011 #3
    Hey, I've edited a portion of my post it wasn't done yet. But I am a bit confused about what you just wrote. Don't you mean:
    [itex]\displaystyle \frac{e^x}{e^x}-\frac{xe^x}{e^x}-\frac{2e^x}{e^x}=1-x-2\,.[/itex]

    Edit: Either way.. thanks!! I will edit my post and make the correction accordingly.
     
  5. Oct 29, 2011 #4
    I don't understand how you got +2. I re-ran my calculation as [tex]=1-x-2[/tex]. But it seems that I only get the right answer if I put [tex]=1-x+2[/tex]

    Could you explain how its +2 and not -2? This is really testing my faith in my algebra foundation right now..
     
  6. Oct 29, 2011 #5

    Deveno

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    [tex]-e^x(x - 2) = -e^x(x) -e^x(-2) = -xe^x + 2e^x[/tex]
     
  7. Oct 29, 2011 #6

    SammyS

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    My earlier post was in error.

    Distributing the ex gives a numerator of:

    ex - xex + 2ex

    Then dividing this by ex gives: 1 - x + 2 = 3 - x .
     
  8. Oct 29, 2011 #7
    0o silly me.. It ended up being a distributive mistake all along 0__O. Hahaha.. I must need a break then.. I find that quite funny. Well thanks for pointing the mistake. :approve:
     
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