What is the extreme f(t)=(3-t)u_{4}(t) ?

The extreme f(t)=(3-t)u_{4}(t) is Saddle(0,0),Saddle(3,0)

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extreme f(t)=(3-t)u_{4}(t)

extreme points

f (t) = (3 - t) u_{4} (t)

Saddle (0, 0), Saddle (3, 0)

Solution steps

Find the critical points:

(0, 0), (3, 0)

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

D (t, u_{4}) = - 4 t^{2} + 12 t - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(3, 0) :

Saddle

Saddle (0, 0), Saddle (3, 0)

x x^{2} + 4

f (x) = 2 x^{3} - 30 x^{2} + 126 x - 3

f (x) = x^{2} + y^{2} - 5 x - 4 y + x y

f (x) = 6 x^{2} - 8 x^{4}

f (x) = 4 x + x^{2} + 2

What is the extreme f(t)=(3-t)u_{4}(t) ?
The extreme f(t)=(3-t)u_{4}(t) is Saddle(0,0),Saddle(3,0)