What is the extreme f(x,y)=x^2+x-3xy+y^3-5 ?

The extreme f(x,y)=x^2+x-3xy+y^3-5 is Minimum(1,1),Saddle(1/4 , 1/2)

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extreme f(x,y)=x^2+x-3xy+y^3-5

extreme points

f (x, y) = x^{2} + x - 3 x y + y^{3} - 5

Minimum (1, 1), Saddle (\frac{1}{4}, \frac{1}{2})

Solution steps

Find the critical points:

(1, 1), (\frac{1}{4}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 12 y - 9

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(\frac{1}{4}, \frac{1}{2}) :

Saddle

Minimum (1, 1), Saddle (\frac{1}{4}, \frac{1}{2})

y = 5 - 2 x^{2}, - 4 \leq x \leq 2

f (x) = x^{3} - \frac{7}{2} x^{2} + 2 x - 5

f (x) = (t^{2} - 4)^{3}

\frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

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

What is the extreme f(x,y)=x^2+x-3xy+y^3-5 ?
The extreme f(x,y)=x^2+x-3xy+y^3-5 is Minimum(1,1),Saddle(1/4 , 1/2)