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

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

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

extreme points

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

Minimum (0, \frac{1}{2})

Solution steps

Find the critical points:

(0, \frac{1}{2})

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, \frac{1}{2}) :

Minimum

Minimum (0, \frac{1}{2})

f (x) = x^{3} + 6 x^{2} + 12 x + 5

f (x) = x^{4} - 98 x^{2} + 4

(x + 3)^{2} (x^{3} - x^{2})

f (x) = (x + 2)^{3} - 5

f (x) = (x - 2) e^{x}

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