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Jusaquikie
Group Title
Find the critical numbers of the function. (Enter your answers as a commaseparated list. If an answer does not exist, enter DNE.)
f(x) = x^−6 ln x
 2 years ago
 2 years ago
Jusaquikie Group Title
Find the critical numbers of the function. (Enter your answers as a commaseparated list. If an answer does not exist, enter DNE.) f(x) = x^−6 ln x
 2 years ago
 2 years ago

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Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
the actual question is \[x^{6}\ln x\] I think the derivative is \[x^{7}6x^{7}\ln x \]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
derivative is better written as \[\frac{16\ln(x)}{x^7}\] makes finding the critical points a lot easier
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
on critical point is 0, but that is not in the domain of your function. so you can ignore it other one solve \[16\ln(x)=0\]
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
i think part of my problem is i don't know how to solve for ln
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[16\ln(x)=0\] \[6\ln(x)=1\] \[\ln(x)=\frac{1}{6}\] \[x=e^{\frac{1}{6}}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
like that you are going to have to "exponentiate" at some point
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[\ln(x)=y\iff e^y=x\]
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
so if i had ln23 that would be e^23?
 one year ago

Jusaquikie Group TitleBest ResponseYou've already chosen the best response.0
actually ln23 is 3.13 so E^3.13 = 23?
 one year ago
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