What is the extreme P(a)=y^2-4r^2+r+e^2 ?

The extreme P(a)=y^2-4r^2+r+e^2 is Saddle(0, 1/8)

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extreme P(a)=y^2-4r^2+r+e^2

extreme points

P (a) = y^{2} - 4 r^{2} + r + e^{2}

Saddle (0, \frac{1}{8})

Solution steps

Find the critical points:

(0, \frac{1}{8})

\frac{\partial ^{2} f}{\partial y ^{2}} = - 8

D (y, r) = - 16

Check the sign of at each critical point

Check critical point

(0, \frac{1}{8}) :

Saddle

Saddle (0, \frac{1}{8})

f (t) = 25 cos (2 t)

f (x) = - 2 x^{3} + 24 x^{2} - 42 x + 7

f (x) = - 0.01 x^{2} + 120 x + 200000

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

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

What is the extreme P(a)=y^2-4r^2+r+e^2 ?
The extreme P(a)=y^2-4r^2+r+e^2 is Saddle(0, 1/8)