What is the extreme g(t)=e^{4t^3}xln(3t-1) ?

The extreme g(t)=e^{4t^3}xln(3t-1) is Saddle(2/3 ,0)

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extreme g(t)=e^{4t^3}xln(3t-1)

extreme points

g (t) = e^{4 t^{3}} x ln (3 t - 1)

Saddle (\frac{2}{3}, 0)

Solution steps

Find the critical points:

(\frac{2}{3}, 0)

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

D (t, x) = - 144 e^{8 t^{3}} t^{4} ln^{2} (3 t - 1) - \frac{72 e ^{8 t^{3}} t ^{2} ln ( 3 t - 1 )}{3 t - 1} - \frac{9 e ^{8 t^{3}}}{( 3 t - 1 ) ^{2}}

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, 0) :

Saddle

Saddle (\frac{2}{3}, 0)

What is the extreme g(t)=e^{4t^3}xln(3t-1) ?
The extreme g(t)=e^{4t^3}xln(3t-1) is Saddle(2/3 ,0)