What is the extreme f(x)=(9x^2-1)(1+4y) ?

The extreme f(x)=(9x^2-1)(1+4y) is Saddle(1/3 ,-1/4),Saddle(-1/3 ,-1/4)

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extreme f(x)=(9x^2-1)(1+4y)

extreme points

f (x) = (9 x^{2} - 1) (1 + 4 y)

Saddle (\frac{1}{3}, - \frac{1}{4}), Saddle (- \frac{1}{3}, - \frac{1}{4})

Solution steps

Find the critical points:

(\frac{1}{3}, - \frac{1}{4}), (- \frac{1}{3}, - \frac{1}{4})

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

D (x, y) = - 5184 x^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, - \frac{1}{4}) :

Saddle

Check critical point

(- \frac{1}{3}, - \frac{1}{4}) :

Saddle

Saddle (\frac{1}{3}, - \frac{1}{4}), Saddle (- \frac{1}{3}, - \frac{1}{4})

f (x) = 2 x^{3} + 6 x - 9

\frac{1}{3} π r^{2} h

f (x) = - 2 x^{2} + 2 x + 4

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

f (x, y) = 10 x^{2} - 4 x y + 25 y^{2}

What is the extreme f(x)=(9x^2-1)(1+4y) ?
The extreme f(x)=(9x^2-1)(1+4y) is Saddle(1/3 ,-1/4),Saddle(-1/3 ,-1/4)