What is the extreme f(x)=x^3+2y^2+3xy+8 ?

The extreme f(x)=x^3+2y^2+3xy+8 is Saddle(0,0),Minimum(3/4 ,-9/16)

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extreme f(x)=x^3+2y^2+3xy+8

extreme points

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

Saddle (0, 0), Minimum (\frac{3}{4}, - \frac{9}{16})

Solution steps

Find the critical points:

(0, 0), (\frac{3}{4}, - \frac{9}{16})

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

D (x, y) = 24 x - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{3}{4}, - \frac{9}{16}) :

Minimum

Saddle (0, 0), Minimum (\frac{3}{4}, - \frac{9}{16})

f (x, y) = x^{2} + x y + y^{2} - 6 x - 9 y

\frac{x + 3}{x + 2}

f (x) = x (\frac{39900 - x}{5})

f (x) = x^{3} - x^{2} - x + 5, - 1 \leq x \leq 2

f (x) = 24 x - 24 x^{2} + 6 x^{3}

What is the extreme f(x)=x^3+2y^2+3xy+8 ?
The extreme f(x)=x^3+2y^2+3xy+8 is Saddle(0,0),Minimum(3/4 ,-9/16)