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mathcalculus

  • 2 years ago

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

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  1. mathcalculus
    • 2 years ago
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    i used the quotient rule!

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

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

  4. anonymous
    • 2 years ago
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    i doubt there are any

  5. mathcalculus
    • 2 years ago
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    why?

  6. anonymous
    • 2 years ago
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    the derivative is \(\frac{-5}{(x-1)^2}\)

  7. anonymous
    • 2 years ago
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    a fraction is only zero if the numerator is zero this numerator is \(-5\)

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

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

  10. anonymous
    • 2 years 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
    • 2 years ago
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    right

  12. anonymous
    • 2 years ago
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    you may remember what such a thing looks like from a pre - calc course

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

  14. mathcalculus
    • 2 years ago
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    i know 1 gives you undefined.

  15. anonymous
    • 2 years 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
    • 2 years ago
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    you mean the denominator?

  17. anonymous
    • 2 years 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
    • 2 years ago
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    i did, but what i don't understand why it has no critical point.

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

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

  21. mathcalculus
    • 2 years 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. anonymous
    • 2 years 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. anonymous
    • 2 years ago
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    when is a fraction equal to zero?

  24. mathcalculus
    • 2 years ago
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    oh no it doesn't.

  25. mathcalculus
    • 2 years ago
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    and 0 does not multiply -5..

  26. anonymous
    • 2 years ago
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    is \(\frac{5}{x}=0\) solvable?

  27. mathcalculus
    • 2 years ago
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    nope.

  28. mathcalculus
    • 2 years ago
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    i think i got it...

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

  30. mathcalculus
    • 2 years ago
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    thanks @satellite73 :)

  31. anonymous
    • 2 years ago
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    how about \(e^x\)? is that ever zero? yw

  32. mathcalculus
    • 2 years ago
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    no

  33. mathcalculus
    • 2 years ago
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    1?

  34. anonymous
    • 2 years ago
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    no, \(e^x>0\) for all \(x\) \(e^1=e\) not zero

  35. anonymous
    • 2 years ago
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    the function \(f(x)=e^x\) is strictly increasing.

  36. mathcalculus
    • 2 years ago
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    right.

  37. mathcalculus
    • 2 years 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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