Find the extreme values of the function and where they occur. g(x)= x^2lnx

Question

zarkon · Answer

find g'
set equal to zero

solve for x

anonymous · Answer

x[2ln(x)+1]=0

zarkon · Answer

so x=0 or [2ln(x)+1]=0

zarkon · Answer

can you solve this?

anonymous · Answer

I need a little help

zarkon · Answer

subtract 1

then divide by 2

zarkon · Answer

[2ln(x)+1]=0
2ln(x)=-1

ln(x)=-1/2

zarkon · Answer

so x is ...

anonymous · Answer

thats actually the part i needed help with. I had ln(x)= -1/2

zarkon · Answer

$$\ln(x)=-1/2$$

$$e^{\ln(x)}=e^{-1/2}$$

$$x=e^{-1/2}$$

anonymous · Answer

So is that an extreme value?

zarkon · Answer

that is a critical number

anonymous · Answer

Alright, so how do I find the extreme values? Sorry I'm not very good at this.

zarkon · Answer

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 · Answer

Ok I got -.183939

zarkon · Answer

ok...now you need to see if it is a max or min...

anonymous · Answer

How do I do the first derivative test?

zarkon · Answer

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 · Answer

I tried plugging in -.5 and 0 but you cant take the ln of 0 or a negative

zarkon · Answer

try .5 and 1

you are picking numbers about the critical number e^(-1/2) which is approx .60653

anonymous · Answer

so .5 is negative and 1 is positive

zarkon · Answer

therefore you have a min

anonymous · Answer

a min at e^-1/2?

zarkon · Answer

yes the min is located  at x=e^{-1/2}

anonymous · Answer

great! thank you