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mathcalculus

  • one year ago

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

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  1. mathcalculus
    • one year ago
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    i used the quotient rule!

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

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

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

  5. mathcalculus
    • one year ago
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    why?

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

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

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

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

  10. satellite73
    • one year ago
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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

  11. mathcalculus
    • one year ago
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    right

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

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

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

  15. satellite73
    • one year ago
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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

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

  17. satellite73
    • one year ago
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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

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

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

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

  21. mathcalculus
    • one year ago
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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

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

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

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

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

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

  27. mathcalculus
    • one year ago
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    nope.

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

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

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

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

  32. mathcalculus
    • one year ago
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    no

  33. mathcalculus
    • one year ago
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    1?

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

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

  36. mathcalculus
    • one year ago
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    right.

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

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