What is the extreme f(y)=6x^2+7y^2 ?

The extreme f(y)=6x^2+7y^2 is Minimum(0,0)

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extreme f(y)=6x^2+7y^2

extreme points

f (y) = 6 x^{2} + 7 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial y ^{2}} = 12

D (y, x) = 168

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{t ^{2}}{1 + t ^{3}}

- x^{3} + 6 x^{2} + x - 1

f (x, y) = x^{3} - 6 x y + y^{3} + 4

- x^{3} + 6 x^{2} + x + 1

\frac{x ^{2} + 1}{( x - 1 ) ^{2}}