1. Originally Posted by db5vry
It isn't, I am not sure how to do that at all. I feel increasingly out of my depth with this question
Clearly, $x^{-\sqrt{x}}\ne0$ so that $f'(x)=0$ precisely when $\frac{1}{\sqrt{x}}+\frac{\ln(x)}{2\sqrt{x}}=\frac{ 2+\ln(x)}{2\sqrt{x}}=0$ which is clearly true when $\ln(x)+2=0\implies \ln(x)=-2\implies x=\tfrac{1}{e^2}$.

2. Originally Posted by Drexel28
Clearly, $x^{-\sqrt{x}}\ne0$ so that $f'(x)=0$ precisely when $\frac{1}{\sqrt{x}}+\frac{\ln(x)}{2\sqrt{x}}=\frac{ 2+\ln(x)}{2\sqrt{x}}=0$ which is clearly true when $\ln(x)+2=0\implies \ln(x)=-2\implies x=\tfrac{1}{e^2}$.
I see! I wasn't thinking of it in that way. I understand that $ln (x)$ is equal to -2, but how did you see that x was equal to $\frac{1}{e^2}$?

3. Originally Posted by db5vry
I see! I wasn't thinking of it in that way. I understand that $ln (x)$ is equal to -2, but how did you see that x was equal to $\frac{1}{e^2}$?
$\ln(x)=-2\implies x=e^{-2}=\frac{1}{e^2}$

4. Originally Posted by Drexel28
$\ln(x)=-2\implies x=e^{-2}=\frac{1}{e^2}$
That's fantastic! I see what you were guiding me towards all along now. I need a break after a couple of hours now and want to thank you for helping me out

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