What is the extreme f(x)=(x^2-4)/(x-1)y ?

The extreme f(x)=(x^2-4)/(x-1)y is Saddle(2,0),Saddle(-2,0)

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extreme f(x)=(x^2-4)/(x-1)y

extreme points

f (x) = \frac{x ^{2} - 4}{x - 1} y

Saddle (2, 0), Saddle (- 2, 0)

Solution steps

Find the critical points:

(2, 0), (- 2, 0)

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

D (x, y) = - \frac{( x ^{2} - 2 x + 4 ) ^{2}}{( x - 1 ) ^{4}}

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Saddle (2, 0), Saddle (- 2, 0)

f (x) = 2 + 12 x^{2} - 8 x^{3}

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

f (x, y, z) = z

- 3^{2 x - 4} - 5

f (x) = (8 x - 9) e^{4 x}

What is the extreme f(x)=(x^2-4)/(x-1)y ?
The extreme f(x)=(x^2-4)/(x-1)y is Saddle(2,0),Saddle(-2,0)