What is the extreme f(x)-x^4-8x^2+1 ?

The extreme f(x)-x^4-8x^2+1 is Saddle(0,0)

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extreme f(x)-x^4-8x^2+1

extreme points

f (x) - x^{4} - 8 x^{2} + 1

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial f ^{2}} = - 12 x^{2} - 16

D (f, x) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = x^{3} - 12 x + 17

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

e^{4 x} + e^{- x}

f (x) = x^{2} - 12 x + 8

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

What is the extreme f(x)-x^4-8x^2+1 ?
The extreme f(x)-x^4-8x^2+1 is Saddle(0,0)