What is the extreme f(t)=(1+2t-3t^2+4t^3)u(t-2) ?

The extreme f(t)=(1+2t-3t^2+4t^3)u(t-2) is Saddle(-0.30448…,0),Saddle(2,0)

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extreme f(t)=(1+2t-3t^2+4t^3)u(t-2)

extreme points

f (t) = (1 + 2 t - 3 t^{2} + 4 t^{3}) u (t - 2)

Saddle (- 0.30448 \dots, 0), Saddle (2, 0)

Solution steps

Find the critical points:

(- 0.30448 \dots, 0), (2, 0)

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

D (t, u) = - (16 t^{3} - 33 t^{2} + 16 t - 3)^{2}

Check the sign of at each critical point

Check critical point

(- 0.30448 \dots, 0) :

Saddle

Check critical point

(2, 0) :

Saddle

Saddle (- 0.30448 \dots, 0), Saddle (2, 0)

What is the extreme f(t)=(1+2t-3t^2+4t^3)u(t-2) ?
The extreme f(t)=(1+2t-3t^2+4t^3)u(t-2) is Saddle(-0.30448…,0),Saddle(2,0)