relative extrema for y=x ln x
did you find the derivative?
need the product rule to find it you gotta find that first
yes is lnx+1
so it is then set it equal to zero, solve for \(x\)
i don't know how to solve it
too many typos \[\ln(x)+1=0\\ \ln(x)=-1\] any idea what comes next?
hint, rewrite in equivalent exponential form
plug it in??
oh no
\[\ln(a)=b\iff a=e^b\]
oh so is x=e^-1
so \[\ln(x)=-1\iff x=e^{-1}=\frac{1}{e}\]
right
now i guess you need to know whether that is a relative max, min, or neither
yea i have to identify that and also find the point of inflection
to identify it you have a choice, but probably the easiest method is to take the second derivative, especially since you have to find it anyways
what is the derivative of \[\ln(x)+1\]?
1/(x+1)
Find the derivative separately.. :)
on no dear, that is the derivative of \(\ln(x+1)\)
the derivative of \(\ln(x)\)is \(\frac{1}{x}\) and the derivative of \(1\) is zero
i thought i need to find the derivative of ln (x+1)
\[y = x \cdot \ln(x) \\ y^{\prime} = \ln(x) + 1 \\ y^{''} = \frac{1}{x}\]
if you evaluate \(\frac{1}{x}\) at \(\frac{1}{e}\) you get \(e\) which is evidently positive that means a) the function is concave up and b) the critical point gives a relative minimum |dw:1413260532792:dw|
oh i get it
and for the point of inflection, i know you have to find the second derivative but i don't know what next
set it equal to zero and solve if you are not thinking straight
if you are thinking straight, then note that the domain of \[f(x)=x\ln(x)\] is \((0,\infty)\) on that interval \(\frac{1}{x}>0\) for any positive \(x\)
meaning the function is always concave up, and there are no inflection points
alright i guess i know how to do it
thank you for your help~
yw
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