What is the extreme f(x)=(x^3-4xy)e^{-2y} ?

The extreme f(x)=(x^3-4xy)e^{-2y} is Saddle(0,0),Minimum(1, 3/4),Maximum(-1, 3/4)

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extreme f(x)=(x^3-4xy)e^{-2y}

extreme points

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

Saddle (0, 0), Minimum (1, \frac{3}{4}), Maximum (- 1, \frac{3}{4})

Solution steps

Find the critical points:

(0, 0), (1, \frac{3}{4}), (- 1, \frac{3}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 x (- 8 e^{- 2 y} - 2 e^{- 2 y} x^{2} + 8 e^{- 2 y} y)

D (x, y) = - 12 e^{- 4 y} x^{4} + 48 e^{- 4 y} x^{2} - 16 e^{- 4 y} - 64 e^{- 4 y} y^{2} + 64 e^{- 4 y} y

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, \frac{3}{4}) :

Minimum

Check critical point

(- 1, \frac{3}{4}) :

Maximum

Saddle (0, 0), Minimum (1, \frac{3}{4}), Maximum (- 1, \frac{3}{4})

f (x) = - 2 (x - 3) (x + 5)

x^{3} - 2 x^{2} - 15 x + 5, - 2 \leq x \leq 0

x^{2} + y x + y^{2}

\frac{7 x - 6}{2 x - 8}

f (x) = 5 x e^{- x}

What is the extreme f(x)=(x^3-4xy)e^{-2y} ?
The extreme f(x)=(x^3-4xy)e^{-2y} is Saddle(0,0),Minimum(1, 3/4),Maximum(-1, 3/4)