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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.

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up vote 1 down vote accepted

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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Oh, I forgot that part, thanks, @DonAntonio – Vaolter Nov 6 '12 at 19:08

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