What is the extreme f(x)=Ix^2-4I ?

The extreme f(x)=Ix^2-4I is Saddle(2,0),Saddle(-2,0)

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extreme f(x)=Ix^2-4I

extreme points

f (x) = I x^{2} - 4 I

Saddle (2, 0), Saddle (- 2, 0)

Solution steps

Find the critical points:

(2, 0), (- 2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, I) = - 4 x^{2}

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Saddle (2, 0), Saddle (- 2, 0)

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

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

f (x) = 8 x ln (x)

f (x) = x^{3} - 3 x^{2} - 9 x + 11

f (x) = x^{3} - 3 x^{2} - 9 x + 20

What is the extreme f(x)=Ix^2-4I ?
The extreme f(x)=Ix^2-4I is Saddle(2,0),Saddle(-2,0)