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mathcalculus Group Title

help! :( finding the critical point of g(x)= (x+4)/(x-1)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    i used the quotient rule!

    • one year ago
  2. mathcalculus Group Title
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    i got my critical point 5 and 1 but it's wrong :(

    • one year ago
  3. mathcalculus Group Title
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    -5/(x^2-2x+1)= 0

    • one year ago
  4. satellite73 Group Title
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    i doubt there are any

    • one year ago
  5. mathcalculus Group Title
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    why?

    • one year ago
  6. satellite73 Group Title
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    the derivative is \(\frac{-5}{(x-1)^2}\)

    • one year ago
  7. satellite73 Group Title
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    a fraction is only zero if the numerator is zero this numerator is \(-5\)

    • one year ago
  8. mathcalculus Group Title
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    -5/(x^2-2x+1)= 0

    • one year ago
  9. mathcalculus Group Title
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    right so when i get to this: -5/(x^2-2x+1)= 0 what do i do?

    • one year ago
  10. satellite73 Group Title
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    the denominator is positive for all values of \(x\) except for \(x=1\) where both it, and the original function are undefined the numerator is negative that tells you the function is always decreasing

    • one year ago
  11. mathcalculus Group Title
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    right

    • one year ago
  12. satellite73 Group Title
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    you may remember what such a thing looks like from a pre - calc course

    • one year ago
  13. mathcalculus Group Title
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    ok but then why isn't -5 considered a critical point?

    • one year ago
  14. mathcalculus Group Title
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    i know 1 gives you undefined.

    • one year ago
  15. satellite73 Group Title
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    a critical point is not the numerator a critical point is a number \(a\) where either \(f'(a)=0\) or \(f'(a)\) does not exist

    • one year ago
  16. mathcalculus Group Title
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    you mean the denominator?

    • one year ago
  17. satellite73 Group Title
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    graph your derivative you will see that it never crosses the \(y\) axis, that it always lives below it i.e. it is always negative if you graph your original function, you will see that it is always decreasing

    • one year ago
  18. mathcalculus Group Title
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    i did, but what i don't understand why it has no critical point.

    • one year ago
  19. mathcalculus Group Title
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    okay but in this question im not allowed to use the calculator.

    • one year ago
  20. satellite73 Group Title
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    well, because it doesn't does \(f(x)=2x+1\) have a critical point?

    • one year ago
  21. mathcalculus Group Title
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    so if i can't graph it, how do can i automatically tell there is no critical point by looking at the derivative -5/(x^2-2x+1)= 0

    • one year ago
  22. satellite73 Group Title
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    of course it doesn't, because for \(f(x)=2x+1\)you have a line that is always increasing

    • one year ago
  23. satellite73 Group Title
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    when is a fraction equal to zero?

    • one year ago
  24. mathcalculus Group Title
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    oh no it doesn't.

    • one year ago
  25. mathcalculus Group Title
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    and 0 does not multiply -5..

    • one year ago
  26. satellite73 Group Title
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    is \(\frac{5}{x}=0\) solvable?

    • one year ago
  27. mathcalculus Group Title
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    nope.

    • one year ago
  28. mathcalculus Group Title
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    i think i got it...

    • one year ago
  29. satellite73 Group Title
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    right, and neither is \(\frac{-5}{(x-1)^2}\)

    • one year ago
  30. mathcalculus Group Title
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    thanks @satellite73 :)

    • one year ago
  31. satellite73 Group Title
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    how about \(e^x\)? is that ever zero? yw

    • one year ago
  32. mathcalculus Group Title
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    no

    • one year ago
  33. mathcalculus Group Title
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    1?

    • one year ago
  34. satellite73 Group Title
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    no, \(e^x>0\) for all \(x\) \(e^1=e\) not zero

    • one year ago
  35. satellite73 Group Title
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    the function \(f(x)=e^x\) is strictly increasing.

    • one year ago
  36. mathcalculus Group Title
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    right.

    • one year ago
  37. mathcalculus Group Title
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    @satellite73 i have a question: how come -5/(x-1)=0 is not a critical point and -4/15 is?

    • one year ago
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