What is the extreme f(x,y)=7+4x^2+2y^2 ?

The extreme f(x,y)=7+4x^2+2y^2 is Minimum(0,0)

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extreme f(x,y)=7+4x^2+2y^2

extreme points

f (x, y) = 7 + 4 x^{2} + 2 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 32

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = - 5 + x - x^{2}

15 y^{30}

f (x) = 3 x^{2} - 30 x + 12, - 10 \leq x \leq 10

\frac{x - 521}{x ^{2} + x - 132}

f (x) = 8 x^{3} - 7 ln (x^{3}), (0.12, 7)

What is the extreme f(x,y)=7+4x^2+2y^2 ?
The extreme f(x,y)=7+4x^2+2y^2 is Minimum(0,0)