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Mathematics 16 Online
OpenStudy (anonymous):

relative extrema for y=x ln x

OpenStudy (anonymous):

did you find the derivative?

OpenStudy (anonymous):

need the product rule to find it you gotta find that first

OpenStudy (anonymous):

yes is lnx+1

OpenStudy (anonymous):

so it is then set it equal to zero, solve for \(x\)

OpenStudy (anonymous):

i don't know how to solve it

OpenStudy (anonymous):

too many typos \[\ln(x)+1=0\\ \ln(x)=-1\] any idea what comes next?

OpenStudy (anonymous):

hint, rewrite in equivalent exponential form

OpenStudy (anonymous):

plug it in??

OpenStudy (anonymous):

oh no

OpenStudy (anonymous):

\[\ln(a)=b\iff a=e^b\]

OpenStudy (anonymous):

oh so is x=e^-1

OpenStudy (anonymous):

so \[\ln(x)=-1\iff x=e^{-1}=\frac{1}{e}\]

OpenStudy (anonymous):

right

OpenStudy (anonymous):

now i guess you need to know whether that is a relative max, min, or neither

OpenStudy (anonymous):

yea i have to identify that and also find the point of inflection

OpenStudy (anonymous):

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

OpenStudy (anonymous):

what is the derivative of \[\ln(x)+1\]?

OpenStudy (anonymous):

1/(x+1)

OpenStudy (anonymous):

Find the derivative separately.. :)

OpenStudy (anonymous):

on no dear, that is the derivative of \(\ln(x+1)\)

OpenStudy (anonymous):

the derivative of \(\ln(x)\)is \(\frac{1}{x}\) and the derivative of \(1\) is zero

OpenStudy (anonymous):

i thought i need to find the derivative of ln (x+1)

OpenStudy (anonymous):

\[y = x \cdot \ln(x) \\ y^{\prime} = \ln(x) + 1 \\ y^{''} = \frac{1}{x}\]

OpenStudy (anonymous):

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|

OpenStudy (anonymous):

oh i get it

OpenStudy (anonymous):

and for the point of inflection, i know you have to find the second derivative but i don't know what next

OpenStudy (anonymous):

set it equal to zero and solve if you are not thinking straight

OpenStudy (anonymous):

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 \(x\)

OpenStudy (anonymous):

meaning the function is always concave up, and there are no inflection points

OpenStudy (anonymous):

alright i guess i know how to do it

OpenStudy (anonymous):

thank you for your help~

OpenStudy (anonymous):

yw

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