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$$ f(x,y) = x(x-1)^2 - 2y^2 $$

the critical points are given by setting the gradient of the function equal to zero. The gradient is given by

$$ \nabla f(x,y) = (x-1)(3x-1)\hat{x} + 4y \hat{y} $$

Setting each component equal to zero, it is clear that ##y = 0## and ## x = 1/3, \, 1##. This therefore gives two critical points: ##C_1 = (1/3, 0)## and ##C_2 = (1,0)##.

The second derivative of the function with respect to ##y## is given by

$$ f_{yy}(x,y) = -4 $$

The second derivative of the function with respect to ##x## is given by

$$ f_{xx}(x,y) = 6x - 4 $$

We can immediately see that ##f_{xx}(1,0) = 2## and ##f_{xx}(1/3,0) = -2## - both non-zero. However, the concavity of ##C_1## is the same in ##x## and ##y##: it is a maximum. The concavities differ at ##C_2##, so it is a saddle point, not an extremum.

Finally, to determine if it is a global extremum, we can simply determine that the value of the function at ##C_1## is ##f(1/3,0)= 4/27##, and show that ##f(\infty,0) = \infty##: clearly, ##C_1## is not a global maximum, and no other extrema exist. Therefore, there are no global extrema.