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mathcalculus

  • 2 years ago

locate the critical points: g(x)= (1*x+1)/1-7x

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  1. mathcalculus
    • 2 years ago
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    so far im left with : 7x^2-7x-6

  2. mathcalculus
    • 2 years ago
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    i know that i have to = to zero

  3. mathcalculus
    • 2 years ago
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    then I'm stuck.

  4. mathcalculus
    • 2 years ago
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    factor?

  5. Peter14
    • 2 years ago
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    could you write out the equation more clearly?

  6. mathcalculus
    • 2 years ago
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    g(x) \[\frac{ 1*x+1 }{1-7x}\]

  7. Peter14
    • 2 years ago
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    then you use quotient rule?

  8. mathcalculus
    • 2 years ago
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    yes

  9. mathcalculus
    • 2 years ago
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    i have 7x^2-7x-6

  10. mattt9
    • 2 years ago
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    |dw:1363323083958:dw|

  11. mathcalculus
    • 2 years ago
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    i found the derivative..

  12. mattt9
    • 2 years ago
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    |dw:1363323153025:dw|

  13. mathcalculus
    • 2 years ago
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    @mattt9 the derivative that i got is this: \[\frac{ 1*(1-7x)-7(x+1) }{ (1-7x)^{2} }\]

  14. mattt9
    • 2 years ago
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    |dw:1363323222354:dw|

  15. mathcalculus
    • 2 years ago
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    how did i get the derivative wrong?

  16. mattt9
    • 2 years ago
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    you missed a negative

  17. Peter14
    • 2 years ago
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    matt, I don't think you can take away the ^-2 like that...

  18. mattt9
    • 2 years ago
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    hmm i think you're right actually.

  19. mathcalculus
    • 2 years ago
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    the answer is none

  20. mattt9
    • 2 years ago
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    okay that's what i thought.

  21. mattt9
    • 2 years ago
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    after peter reclarified that for me

  22. mathcalculus
    • 2 years ago
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    x cannot be 1/7 because you cancel the -7x+7x and youre left with -6

  23. Peter14
    • 2 years ago
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    graphing\[y=\frac{ 1 }{ (1-7x)^2 }\] with Desmos I don't see a place where it crosses the X-axis

  24. mathcalculus
    • 2 years ago
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    when you multiply the denominator to zero= 0

  25. mathcalculus
    • 2 years ago
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    and -6 does not equal to an x

  26. Peter14
    • 2 years ago
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    thinking about it: 1/anything cannot equal 0

  27. Peter14
    • 2 years ago
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    so there are no critical points

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