What is the extreme f(t)=(t)+2u(t)-3e^{-2t}u(t) ?

The extreme f(t)=(t)+2u(t)-3e^{-2t}u(t) is Saddle(0,1),Saddle(-1/2 ln(2/3), 1/(2ln(2/3)))

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extreme f(t)=(t)+2u(t)-3e^{-2t}u(t)

extreme

Saddle (0, 1), Saddle (- \frac{1}{2} ln (\frac{2}{3}), \frac{1}{2 ln ( \frac{2}{3} )})

Solution steps

Find the critical points:

(0, 1), (- \frac{1}{2} ln (\frac{2}{3}), \frac{1}{2 ln ( \frac{2}{3} )})

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

D (t, u) = - 9 e^{- 4 t} (2 t - 1)^{2} - 24 e^{- 2 t} t + 12 e^{- 2 t} - 4

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(- \frac{1}{2} ln (\frac{2}{3}), \frac{1}{2 ln ( \frac{2}{3} )}) :

Saddle

Saddle (0, 1), Saddle (- \frac{1}{2} ln (\frac{2}{3}), \frac{1}{2 ln ( \frac{2}{3} )})

What is the extreme f(t)=(t)+2u(t)-3e^{-2t}u(t) ?
The extreme f(t)=(t)+2u(t)-3e^{-2t}u(t) is Saddle(0,1),Saddle(-1/2 ln(2/3), 1/(2ln(2/3)))