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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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)

f (x) = - 4 x^{3} + 6 x^{2} + 1, - 4 \leq x \leq 3

f (x) = x^{3} - 12 x^{2} + 45 x + 9

f (x, y) = x^{- 1} - 64 y^{- 1} + x y

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

x^{2} - y^{2} - 2 x + 4 y + 6

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)