What is the extreme f(x)=(2x-3)/(x^2-3x)se ?

The extreme f(x)=(2x-3)/(x^2-3x)se is Saddle(3/2 ,0)

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extreme f(x)=(2x-3)/(x^2-3x)se

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, s) = - \frac{e ^{2} ( - 2 x ^{2} + 6 x - 9 ) ^{2}}{( x ^{2} - 3 x ) ^{4}}

Check the sign of at each critical point

Check critical point

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

Saddle

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

f (x, y) = x + \frac{y}{- x + 3 y}

f (x, y) = 9 x^{2} y - 9 x y^{3} + 9 y^{3}

f (x, y) = 3 x^{2} - 5 x y + 6 y^{2} + x - y

f (x) = \frac{( x - 1 ) ^{2} + ( y + 2 ) ^{2}}{2}

f (x) = - 7 (2 x^{3} - \frac{125}{32})^{2} + 1

What is the extreme f(x)=(2x-3)/(x^2-3x)se ?
The extreme f(x)=(2x-3)/(x^2-3x)se is Saddle(3/2 ,0)