What is the extreme f(t)=t+(t^2+t)*u(t-3)+(t-t^2)u(t-5) ?

The extreme f(t)=t+(t^2+t)*u(t-3)+(t-t^2)u(t-5) is Saddle(0, 1/8),Saddle(2,-1/8)

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extreme f(t)=t+(t^2+t)*u(t-3)+(t-t^2)u(t-5)

extreme points

f (t) = t + (t^{2} + t) \cdot u (t - 3) + (t - t^{2}) u (t - 5)

Saddle (0, \frac{1}{8}), Saddle (2, - \frac{1}{8})

Solution steps

Find the critical points:

(0, \frac{1}{8}), (2, - \frac{1}{8})

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

D (t, u) = - 64 t^{2} + 128 t - 64

Check the sign of at each critical point

Check critical point

(0, \frac{1}{8}) :

Saddle

Check critical point

(2, - \frac{1}{8}) :

Saddle

Saddle (0, \frac{1}{8}), Saddle (2, - \frac{1}{8})

2 x^{2} y - 4 y^{2} - 8 y - 4

f (x) = (x + 1)^{3} (x - 1)

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

f (x) = 14 x^{3} + 336 x^{2}

f (x) = x^{2} + \frac{14}{x}

What is the extreme f(t)=t+(t^2+t)*u(t-3)+(t-t^2)u(t-5) ?
The extreme f(t)=t+(t^2+t)*u(t-3)+(t-t^2)u(t-5) is Saddle(0, 1/8),Saddle(2,-1/8)