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Jusaquikie

  • 2 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x) = x^−6 ln x

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  1. Jusaquikie
    • 2 years ago
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    the actual question is \[x^{-6}\ln x\] I think the derivative is \[x^{-7}-6x^{-7}\ln x \]

  2. satellite73
    • 2 years ago
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    derivative is better written as \[\frac{1-6\ln(x)}{x^7}\] makes finding the critical points a lot easier

  3. satellite73
    • 2 years ago
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    on critical point is 0, but that is not in the domain of your function. so you can ignore it other one solve \[1-6\ln(x)=0\]

  4. Jusaquikie
    • 2 years ago
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    i think part of my problem is i don't know how to solve for ln

  5. satellite73
    • 2 years ago
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    \[1-6\ln(x)=0\] \[6\ln(x)=1\] \[\ln(x)=\frac{1}{6}\] \[x=e^{\frac{1}{6}}\]

  6. satellite73
    • 2 years ago
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    like that you are going to have to "exponentiate" at some point

  7. Jusaquikie
    • 2 years ago
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    ok thanks, i know eulers number is related to natural log i just don't know how to convert back and forth. thanks for the help

  8. satellite73
    • 2 years ago
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    \[\ln(x)=y\iff e^y=x\]

  9. Jusaquikie
    • 2 years ago
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    so if i had ln23 that would be e^23?

  10. Jusaquikie
    • 2 years ago
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    actually ln23 is 3.13 so E^3.13 = 23?

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