What is the extreme f(x,y)=(y^2-x)/(x^2+1) ?

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

Popular >

extreme f(x,y)=(y^2-x)/(x^2+1)

extreme points

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

Minimum (1, 0), Saddle (- 1, 0)

Solution steps

Find the critical points:

(1, 0), (- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2}{x ^{2} + 1}

D (x, y) = \frac{- 4 x ^{3} - 4 x ^{2} y ^{2} + 12 x - 4 y ^{2}}{( x ^{2} + 1 ) ^{4}}

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Check critical point

(- 1, 0) :

Saddle

Minimum (1, 0), Saddle (- 1, 0)

f (x) = \frac{8 x}{x ^{2} + 1}

f (x) = \frac{1}{4} x^{4} + x^{3} - 2

f (x) = (\frac{1}{x}) + 2 + (\frac{4}{1 - x})

f (x) = 2 x^{3} - x^{2} - 20 x + 10

f (x) = x^{3} \cdot e^{- 2 x^{2}}

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