What is the extreme f(x)=x^2+y^2-3y ?

The extreme f(x)=x^2+y^2-3y is Minimum(0, 3/2)

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extreme f(x)=x^2+y^2-3y

extreme points

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

Minimum (0, \frac{3}{2})

Solution steps

Find the critical points:

(0, \frac{3}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, \frac{3}{2}) :

Minimum

Minimum (0, \frac{3}{2})

f (x) = x^{5} - 10 x^{4} - 1 = x^{4} (x - 10) - 1

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 4, - 2 \leq x \leq 5

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

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

What is the extreme f(x)=x^2+y^2-3y ?
The extreme f(x)=x^2+y^2-3y is Minimum(0, 3/2)