What is the extreme f(t)=u^1(t)+2u^3(t)-6u^4(t) ?

The extreme f(t)=u^1(t)+2u^3(t)-6u^4(t) is Saddle(0,0),Saddle(0,0.68673…)

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extreme f(t)=u^1(t)+2u^3(t)-6u^4(t)

extreme points

f (t) = u^{1} (t) + 2 u^{3} (t) - 6 u^{4} (t)

Saddle (0, 0), Saddle (0, 0.68673 \dots)

Solution steps

Find the critical points:

(0, 0), (0, 0.68673 \dots)

\frac{\partial ^{2} f}{\partial t ^{2}} = 12 t u - 72 t u^{2}

D (t, u) = - (- 24 u^{3} + 6 u^{2} + 1)^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, 0.68673 \dots) :

Saddle

Saddle (0, 0), Saddle (0, 0.68673 \dots)

f (x) = 5 - 2 x^{2} + 3 x^{3} - x^{4}

f (x, y) = 2 + 9 - x^{2} - y^{2}

f (x) = x^{2} - 5 x + 6

f (x) = x x + 2

f (x) = 4 x y - x^{4} - y^{4}

What is the extreme f(t)=u^1(t)+2u^3(t)-6u^4(t) ?
The extreme f(t)=u^1(t)+2u^3(t)-6u^4(t) is Saddle(0,0),Saddle(0,0.68673…)