What is the extreme f(t)=5-6u(t-2)+3u(t-5) ?

The extreme f(t)=5-6u(t-2)+3u(t-5) is Saddle(-1,0)

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extreme f(t)=5-6u(t-2)+3u(t-5)

extreme points

f (t) = 5 - 6 u (t - 2) + 3 u (t - 5)

Saddle (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

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

D (t, u) = - 9

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Saddle (- 1, 0)

h (t, v) = - 16 t^{2} + v t + k

f (x, y) = 4 x - 8 x y + 2 y + 1

+ 5^{x} - 4

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

x 6 - x

What is the extreme f(t)=5-6u(t-2)+3u(t-5) ?
The extreme f(t)=5-6u(t-2)+3u(t-5) is Saddle(-1,0)