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

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

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extreme f(x,y)= 1/2 (x^3-x-y^2+3)

extreme points

f (x, y) = \frac{1}{2} (x^{3} - x - y^{2} + 3)

Saddle (\frac{1}{3}, 0), Maximum (- \frac{1}{3}, 0)

Solution steps

Find the critical points:

(\frac{1}{3}, 0), (- \frac{1}{3}, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 1

D (x, y) = - 3 x

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, 0) :

Saddle

Check critical point

(- \frac{1}{3}, 0) :

Maximum

Saddle (\frac{1}{3}, 0), Maximum (- \frac{1}{3}, 0)

f (x) = \frac{( 5 x )}{( x ^{2} + 5 )}

f (x) = ln (x + 2) + \frac{4}{x + 2}

f (x) = 4 x^{7}

f (x) = x^{6} + \frac{x}{6}

y = - 57.17 x^{2} + 621.83 x + 1089

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