What is the extreme f(x)=x^2+2y^2+3x+10 ?

The extreme f(x)=x^2+2y^2+3x+10 is Minimum(-3/2 ,0)

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extreme f(x)=x^2+2y^2+3x+10

extreme points

f (x) = x^{2} + 2 y^{2} + 3 x + 10

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

Solution steps

Find the critical points:

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

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x) = 8 cos (θ) + 7 sin (θ)

(11 - x) (x + 1)^{2}

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

f (x) = \frac{7}{2} x^{2} + \frac{7}{2} - 7 x - 4

\frac{2 ln ( x )}{x}

What is the extreme f(x)=x^2+2y^2+3x+10 ?
The extreme f(x)=x^2+2y^2+3x+10 is Minimum(-3/2 ,0)