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mathcalculus Group Title

help: find the critical points: g(x) x+8/x-1

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    \[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]

    • one year ago
  2. mathcalculus Group Title
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    \[\frac{ 1x-1-1x-8 }{ x ^{2}-2x+1 }\]

    • one year ago
  3. mathcalculus Group Title
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    \[\frac{ -9 }{ x ^{2} -2x+1 }\]

    • one year ago
  4. mathcalculus Group Title
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    then I'm stuck here :/

    • one year ago
  5. Twis7ed Group Title
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    Find x in: \[\frac{ -9 }{ x^2-2x+1 } = 0\]

    • one year ago
  6. Twis7ed Group Title
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    Remember that the value of the derivative is the slope of the line of g(x) at that point so when the first derivative is equal to zero, the slope of the curve is zero and it is therefore a critical point.

    • one year ago
  7. mathcalculus Group Title
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    yes so x=0?

    • one year ago
  8. mathcalculus Group Title
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    that's what I did later..

    • one year ago
  9. Twis7ed Group Title
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    No, when x = 0 then the function is equal to -9...find the value of x so that the dericative is equal to 0.

    • one year ago
  10. Twis7ed Group Title
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    Also, since there is no solution for that equation (just realised :p) You can find at which point it is undefined.

    • one year ago
  11. mathcalculus Group Title
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    yes, i end up with -9=0 can you show me?

    • one year ago
  12. mathcalculus Group Title
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    it's undefined right? because the x is not there..

    • one year ago
  13. Twis7ed Group Title
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    What do you mean?

    • one year ago
  14. mathcalculus Group Title
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    unless it was something like -9x/ x^2-2x+1

    • one year ago
  15. mathcalculus Group Title
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    then there would be an x.

    • one year ago
  16. mathcalculus Group Title
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    but since -9 is alone... there is no critical point. right?

    • one year ago
  17. Twis7ed Group Title
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    Correct, but there IS an asymptote in the graph so I'm not sure if you'd consider that a critical point in your class?

    • one year ago
  18. mathcalculus Group Title
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    whats the asymptote?

    • one year ago
  19. Twis7ed Group Title
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    At x=1

    • one year ago
  20. mathcalculus Group Title
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    how did you find 1 again?

    • one year ago
  21. mathcalculus Group Title
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    btw thank you

    • one year ago
  22. Twis7ed Group Title
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    Set the bottom of the function equal to zero so x^2-2x+1 = 0 and solve for x. No problem :D

    • one year ago
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