True of false about partial derivative

zhuyilun
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Homework Statement


if (2,1) is a critical point of f and fxx(2,1)fyy(2,1) < (fxy(2,1))^2
then f has a saddle point at (2,1)


Homework Equations





The Attempt at a Solution


i think its right
but it turns out to be wrong
can someone tell me why?
 
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wouldn't it need fx(2,1) = fy(2,1) = 0 as well?
 
lanedance said:
wouldn't it need fx(2,1) = fy(2,1) = 0 as well?

could you please explain it a little bit more?
 
ok now i see it is already a critical point, i would agree with you then
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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