I am trying to determine the maxima and minima of $y = x^{1-x}$. So that means looking for the critical points so I have $y^\prime = \frac{1-x}{x}-\log x$. From this $x=0$ is a critical point. Now I want to equate $y^\prime$ to $0$ but the resulting equation in nonlinear. How do I proceed? The question arises where I should just reason it out without calculating aids.
Investigating maxima and minima for $x^{1-x}$
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calculus
2 Answers
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What you have given is actually $ \left(\log(y)\right)'=\frac{y'}{y}=\frac{1-x}{x}-\log(x)\tag{1} $ but it gives the same critical points (when $y\ne0$).
In this case, the solution to $(1)$ can be gotten by inspection to be $x=1$.
However, in general, the trick is to get the equation into a form for the Lambert-W function: $ \frac1x+\log\left(\frac1x\right)=1\Rightarrow\frac1xe^{1/x}=e^1\tag{2} $ Equation $(2)$ says that $ \frac1x=W(e)\Rightarrow x=\frac1{W(e)}=1\tag{3} $
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$f(x):=x^{1-x}=e^{(1-x)\log x}\Longrightarrow f'(x)=\left(-\log x+\frac{1-x}{x}\right)e^{(1-x)\log x}=$
$=-x^{1-x}\log x+(1-x)x^{-x}=x^{-x}\left(-x\log x+1-x\right)$
Check your derivative.
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1Oh, I forgot that part, thanks, @DonAntonio – 2012-11-06