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

The extreme f(x,y)=yln(x)+xy^2 is Minimum(e^2,-1/(e^2)),Saddle(1,0)

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

extreme points

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

Minimum (e^{2}, - \frac{1}{e ^{2}}), Saddle (1, 0)

Solution steps

Find the critical points:

(e^{2}, - \frac{1}{e ^{2}}), (1, 0)

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

D (x, y) = \frac{- 4 x ^{2} y ^{2} - 6 x y - 1}{x ^{2}}

Check the sign of at each critical point

Check critical point

(e^{2}, - \frac{1}{e ^{2}}) :

Minimum

Check critical point

(1, 0) :

Saddle

Minimum (e^{2}, - \frac{1}{e ^{2}}), Saddle (1, 0)

\frac{x}{x ^{2} - 1}

f (x) = \frac{( x ^{2} - 4 )}{( x ^{2} + 4 )}

f (x) = x \cdot ln (x)

f (x, y) = 2 x^{2} + 3 x y + y^{2} + 0 x + 5

f (x, y) = y^{2} (x - 1)^{2} + x^{2} (y + 4)^{2}

What is the extreme f(x,y)=yln(x)+xy^2 ?
The extreme f(x,y)=yln(x)+xy^2 is Minimum(e^2,-1/(e^2)),Saddle(1,0)