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mathcalculus

  • one year ago

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

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  1. mathcalculus
    • one year ago
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    \[\frac{ d }{ dx} \frac{ x+8 }{ x-1 }= \frac{ 1(x-1)- 1(x+8)}{(x-1)(x-1) }\]

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

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

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

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

  6. Twis7ed
    • one year ago
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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.

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

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

  9. Twis7ed
    • one year ago
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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.

  10. Twis7ed
    • one year ago
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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.

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

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

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

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

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

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

  17. Twis7ed
    • one year ago
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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?

  18. mathcalculus
    • one year ago
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    whats the asymptote?

  19. Twis7ed
    • one year ago
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    At x=1

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

  21. mathcalculus
    • one year ago
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    btw thank you

  22. Twis7ed
    • one year ago
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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

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