How to Find Critical Points of f(x) = xln(x)

jtulloss
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I need help finding the critical points of this equation:

f(x) = xln(x)

I found f'(x) to be ln(x)+x, but I don't know how to solve for x when setting f'(x) to 0 to find the critical points. I know I can find the zeros of f'(x) graphically on a calculator, but I need to know how to do it algebraically.

Thanks.
 
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jtulloss said:
I need help finding the critical points of this equation:

f(x) = xln(x)

I found f'(x) to be ln(x)+x,
It's not lnx + x.
 
Ah man, I can't believe I overlooked that. I've been doing that a lot lately with studying for exams and all. I got it now.

Thanks
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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