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

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

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

extreme points

f (x, y) = 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) = - 7 x^{3} + 21 x + 4

x^{3} - 2 x^{2} - 4 x + 4

f (x) = 16 x + 17 x^{\frac{16}{17}}

f (x) = x^{\frac{5}{7}}

y = 9 sin (∣ x ∣), - 2 π \leq x \leq 2 π

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