anonymous
  • anonymous
Find the extreme values of the function and where they occur. g(x)= x^2lnx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Zarkon
  • Zarkon
find g' set equal to zero solve for x
anonymous
  • anonymous
x[2ln(x)+1]=0
Zarkon
  • Zarkon
so x=0 or [2ln(x)+1]=0

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Zarkon
  • Zarkon
can you solve this?
anonymous
  • anonymous
I need a little help
Zarkon
  • Zarkon
subtract 1 then divide by 2
Zarkon
  • Zarkon
[2ln(x)+1]=0 2ln(x)=-1 ln(x)=-1/2
Zarkon
  • Zarkon
so x is ...
anonymous
  • anonymous
thats actually the part i needed help with. I had ln(x)= -1/2
Zarkon
  • Zarkon
\[\ln(x)=-1/2\] \[e^{\ln(x)}=e^{-1/2}\] \[x=e^{-1/2}\]
anonymous
  • anonymous
So is that an extreme value?
Zarkon
  • Zarkon
that is a critical number
anonymous
  • anonymous
Alright, so how do I find the extreme values? Sorry I'm not very good at this.
Zarkon
  • Zarkon
plug the critical number back into the original function to get the value...to test if it is a min/max/ or neither you need to use the first or second derivative test.
anonymous
  • anonymous
Ok I got -.183939
Zarkon
  • Zarkon
ok...now you need to see if it is a max or min...
anonymous
  • anonymous
How do I do the first derivative test?
Zarkon
  • Zarkon
plug numbers that are close to you critical number (above and below) into the derivative to see if the function is increasing or decreasing.
anonymous
  • anonymous
I tried plugging in -.5 and 0 but you cant take the ln of 0 or a negative
Zarkon
  • Zarkon
try .5 and 1 you are picking numbers about the critical number e^(-1/2) which is approx .60653
anonymous
  • anonymous
so .5 is negative and 1 is positive
Zarkon
  • Zarkon
therefore you have a min
anonymous
  • anonymous
a min at e^-1/2?
Zarkon
  • Zarkon
yes the min is located at x=e^{-1/2}
anonymous
  • anonymous
great! thank you

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