What is the extreme f(t)=tu(t-3) ?

The extreme f(t)=tu(t-3) is Saddle(0,0),Saddle(3,0)

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extreme f(t)=tu(t-3)

extreme points

f (t) = t u (t - 3)

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 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)

f (x) = 6 x - 12

2 x^{5} + 5 x^{4}, - 2 \leq x \leq 1

g (x) = \frac{2}{3} x^{3} - 2 x^{2}

f (x) = 6 x^{3} - 9 x^{2} - 216 x + 7

\frac{x}{x ^{2} + 64}

What is the extreme f(t)=tu(t-3) ?
The extreme f(t)=tu(t-3) is Saddle(0,0),Saddle(3,0)