What is the extreme f(x)=(x-1)^2+2y^2 ?

The extreme f(x)=(x-1)^2+2y^2 is Minimum(1,0)

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extreme f(x)=(x-1)^2+2y^2

extreme points

f (x) = (x - 1)^{2} + 2 y^{2}

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

f (x) = - x^{4} + 2 x^{3} + 8 x

f (x) = - x^{3}

f (x) = x^{2} + 10

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 6 x + 7 y

f (x) = 3 x^{4} - 16 x^{3} + 18 x^{2}, - 1 \leq x \leq 4

What is the extreme f(x)=(x-1)^2+2y^2 ?
The extreme f(x)=(x-1)^2+2y^2 is Minimum(1,0)