What is the extreme f(x,y)=x^3-y^3+43xy-7 ?

The extreme f(x,y)=x^3-y^3+43xy-7 is Saddle(0,0),Minimum(43/3 ,-43/3)

Popular >

extreme f(x,y)=x^3-y^3+43xy-7

extreme points

f (x, y) = x^{3} - y^{3} + 43 x y - 7

Saddle (0, 0), Minimum (\frac{43}{3}, - \frac{43}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{43}{3}, - \frac{43}{3})

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

D (x, y) = - 36 x y - 1849

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{43}{3}, - \frac{43}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{43}{3}, - \frac{43}{3})

f (x, y) = 2^{2} \cdot x^{4} + 2^{2} \cdot y^{4} - 4 \cdot x \cdot y + 13

f (x) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

f (x) = \frac{( ln ( x ) ) ^{2}}{x}

f (x) = \frac{x ^{4}}{4} - x^{3}

f (x) = - 4 x^{2} - 16 x

What is the extreme f(x,y)=x^3-y^3+43xy-7 ?
The extreme f(x,y)=x^3-y^3+43xy-7 is Saddle(0,0),Minimum(43/3 ,-43/3)