$$ y=\frac{\ln^2x}{x}$$
finally I got
$t = 0$ and $t = -2$ after $\ln x = t$
but I am not sure if it is correct cause both of them do not exist by default
find maxima and minima of $ y=\frac{\ln^2x}{x}$
0
$\begingroup$
functions
derivatives
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1What is $\ln^2x$? Is that $(\ln x)^2$ or $\ln(\ln(x))\,$? – 2017-02-20
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0Assuming this is $y=(\ln x)^2 / x$ for $x\gt 0$, there is a global minimum and a local maximum – 2017-02-20
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0To differentiate your function, use the chain rule on the numerator and then use the quotient rule. – 2017-02-20
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0They do exist;$\ln(0),\ln(-2)$ do not exist but $\ln x=0\implies x=1$ and $\ln x=-2\implies x=e^{-2}$ – 2017-02-20
2 Answers
1
Hint:
we have: $$ y'=\frac{2\ln x-\ln^2 x}{x^2} $$ so $y'=0$ for: $$ \ln x=0 \quad \mbox{or} \quad \ln x=2 $$
can you solve ?
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If $f(x)=\dfrac{\ln^2(x)}{x^2}$ then $f'(x)=\frac{\ln x(\ln x-2)}{x^2}$. So $\ln x=0$, which means $x=1$, or $\ln x=2$, which means $x=e^2$.
I'll let you check the second derivative to know which one is a maximum and which one is a minimum.