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mathcalculus

  • 2 years ago

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

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

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

  4. mathcalculus
    • 2 years ago
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    then I'm stuck here :/

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

  6. Twis7ed
    • 2 years 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
    • 2 years ago
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    yes so x=0?

  8. mathcalculus
    • 2 years ago
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    that's what I did later..

  9. Twis7ed
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    yes, i end up with -9=0 can you show me?

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

  13. Twis7ed
    • 2 years ago
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    What do you mean?

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

  15. mathcalculus
    • 2 years ago
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    then there would be an x.

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

  17. Twis7ed
    • 2 years 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
    • 2 years ago
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    whats the asymptote?

  19. Twis7ed
    • 2 years ago
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    At x=1

  20. mathcalculus
    • 2 years ago
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    how did you find 1 again?

  21. mathcalculus
    • 2 years ago
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    btw thank you

  22. Twis7ed
    • 2 years 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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