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

The extreme f(t)=t^2u(t)-(3t-2)u(t-2) is Saddle(0.61170…,0)

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extreme f(t)=t^2u(t)-(3t-2)u(t-2)

extreme points

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

Saddle (0.61170 \dots, 0)

Solution steps

Find the critical points:

(0.61170 \dots, 0)

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

D (t, u) = - 9 t^{4} + 36 t^{3} - 84 t^{2} + 96 t - 64

Check the sign of at each critical point

Check critical point

(0.61170 \dots, 0) :

Saddle

Saddle (0.61170 \dots, 0)

f (x) = x^{\frac{4}{5}} (x + 1)

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

f (x, y) = - x^{2} - 3 y^{2} - 2 x y + 4 x - 3

f (x, y) = 13 x e^{y}

y = - x^{2} + 4

What is the extreme f(t)=t^2u(t)-(3t-2)u(t-2) ?
The extreme f(t)=t^2u(t)-(3t-2)u(t-2) is Saddle(0.61170…,0)