What is the extreme f(x,y)=x^2+3xy+2y^4 ?

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

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extreme f(x,y)=x^2+3xy+2y^4

extreme points

f (x, y) = x^{2} + 3 x y + 2 y^{4}

Saddle (0, 0), Minimum (\frac{9}{8}, - \frac{3}{2 ^{2}}), Minimum (- \frac{9}{8}, \frac{3}{2 ^{2}})

Solution steps

Find the critical points:

(0, 0), (\frac{9}{8}, - \frac{3}{2 ^{2}}), (- \frac{9}{8}, \frac{3}{2 ^{2}})

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y^{2}

D (x, y) = 48 y^{2} - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{9}{8}, - \frac{3}{2 ^{2}}) :

Minimum

Check critical point

(- \frac{9}{8}, \frac{3}{2 ^{2}}) :

Minimum

Saddle (0, 0), Minimum (\frac{9}{8}, - \frac{3}{2 ^{2}}), Minimum (- \frac{9}{8}, \frac{3}{2 ^{2}})

f (x) = x^{2} + 6 x + 15

f (x) = ln (x)

f (x) = - x^{3} - 9 x^{2}

f (x) = - 3

f (x) = 3 sin^{2} (x)

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