What is the extreme f(x)=(8x^2-10x-1)/(2x-3)y ?

The extreme f(x)=(8x^2-10x-1)/(2x-3)y is Saddle((5+sqrt(33))/8 ,0),Saddle((5-sqrt(33))/8 ,0)

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extreme f(x)=(8x^2-10x-1)/(2x-3)y

extreme points

f (x) = \frac{8 x ^{2} - 10 x - 1}{2 x - 3} y

Saddle (\frac{5 + 33}{8}, 0), Saddle (\frac{5 - 33}{8}, 0)

Solution steps

Find the critical points:

(\frac{5 + 33}{8}, 0), (\frac{5 - 33}{8}, 0)

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

D (x, y) = - \frac{( 16 x ^{2} - 48 x + 32 ) ^{2}}{( 2 x - 3 ) ^{4}}

Check the sign of at each critical point

Check critical point

(\frac{5 + 33}{8}, 0) :

Saddle

Check critical point

(\frac{5 - 33}{8}, 0) :

Saddle

Saddle (\frac{5 + 33}{8}, 0), Saddle (\frac{5 - 33}{8}, 0)

f (x) = \frac{1}{3} x^{3} - 5 x, - 8 \leq x \leq 8

y = ln (x^{2} + 2)

f (x) = e^{2 x} (x^{2} + 2 x)

f (x, y) = 2 x^{2} - 4 x + 3 y^{2} + 7

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

What is the extreme f(x)=(8x^2-10x-1)/(2x-3)y ?
The extreme f(x)=(8x^2-10x-1)/(2x-3)y is Saddle((5+sqrt(33))/8 ,0),Saddle((5-sqrt(33))/8 ,0)