What is the extreme g(t)=u(t-2)(t^2+2t-1) ?

The extreme g(t)=u(t-2)(t^2+2t-1) is Saddle(2,0),Saddle(-1+sqrt(2),0),Saddle(-1-sqrt(2),0)

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extreme g(t)=u(t-2)(t^2+2t-1)

extreme points

g (t) = u (t - 2) (t^{2} + 2 t - 1)

Saddle (2, 0), Saddle (- 1 + 2, 0), Saddle (- 1 - 2, 0)

Solution steps

Find the critical points:

(2, 0), (- 1 + 2, 0), (- 1 - 2, 0)

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

D (t, u) = - 9 t^{4} + 30 t^{2} - 25

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(- 1 + 2, 0) :

Saddle

Check critical point

(- 1 - 2, 0) :

Saddle

Saddle (2, 0), Saddle (- 1 + 2, 0), Saddle (- 1 - 2, 0)

3 - 2 x

f (x, y) = x^{3} + y^{3} + 3 x y^{2} - 18 (x y)

2 cos (2 x)

0.01 x^{3} - 0.45 x^{2} + 2.43 x + 300

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

What is the extreme g(t)=u(t-2)(t^2+2t-1) ?
The extreme g(t)=u(t-2)(t^2+2t-1) is Saddle(2,0),Saddle(-1+sqrt(2),0),Saddle(-1-sqrt(2),0)