What is the extreme f(x)=2-x^4+2x^2-y^2 ?

The extreme f(x)=2-x^4+2x^2-y^2 is Saddle(0,0),Maximum(-1,0),Maximum(1,0)

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extreme f(x)=2-x^4+2x^2-y^2

extreme points

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

Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0)

Solution steps

Find the critical points:

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

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

D (x, y) = - 2 (- 12 x^{2} + 4)

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 1, 0) :

Maximum

Check critical point

(1, 0) :

Maximum

Saddle (0, 0), Maximum (- 1, 0), Maximum (1, 0)

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

f (x) = \frac{1 + x}{6 - x}

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

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

\frac{x ^{2} - 7 x + 26}{x - 5}

What is the extreme f(x)=2-x^4+2x^2-y^2 ?
The extreme f(x)=2-x^4+2x^2-y^2 is Saddle(0,0),Maximum(-1,0),Maximum(1,0)