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mathcalculus Group Title

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

  • 9 months ago
  • 9 months ago

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  1. mathcalculus Group Title
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    i used the quotient rule!

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

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

    • 9 months ago
  4. satellite73 Group Title
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    i doubt there are any

    • 9 months ago
  5. mathcalculus Group Title
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    why?

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

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

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

    • 9 months 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?

    • 9 months 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

    • 9 months ago
  11. mathcalculus Group Title
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    right

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

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

    • 9 months ago
  14. mathcalculus Group Title
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    i know 1 gives you undefined.

    • 9 months 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

    • 9 months ago
  16. mathcalculus Group Title
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    you mean the denominator?

    • 9 months 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

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

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

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

    • 9 months 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

    • 9 months 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

    • 9 months ago
  23. satellite73 Group Title
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    when is a fraction equal to zero?

    • 9 months ago
  24. mathcalculus Group Title
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    oh no it doesn't.

    • 9 months ago
  25. mathcalculus Group Title
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    and 0 does not multiply -5..

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

    • 9 months ago
  27. mathcalculus Group Title
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    nope.

    • 9 months ago
  28. mathcalculus Group Title
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    i think i got it...

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

    • 9 months ago
  30. mathcalculus Group Title
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    thanks @satellite73 :)

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

    • 9 months ago
  32. mathcalculus Group Title
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    no

    • 9 months ago
  33. mathcalculus Group Title
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    1?

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

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

    • 9 months ago
  36. mathcalculus Group Title
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    right.

    • 9 months 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?

    • 9 months ago
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