What is the extreme f(x)=x-1/2*y^2-1/3 x^3+y+6 ?

The extreme f(x)=x-1/2*y^2-1/3 x^3+y+6 is Saddle(-1,1),Maximum(1,1)

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extreme f(x)=x-1/2*y^2-1/3 x^3+y+6

extreme points

f (x) = x - \frac{1}{2} \cdot y^{2} - \frac{1}{3} x^{3} + y + 6

Saddle (- 1, 1), Maximum (1, 1)

Solution steps

Find the critical points:

(- 1, 1), (1, 1)

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

D (x, y) = 2 x

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Saddle

Check critical point

(1, 1) :

Maximum

Saddle (- 1, 1), Maximum (1, 1)

f (x) = x^{4} - 8 x^{3} + 18 x^{2}

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (- 1, 10)

y = 2 x^{3} - 24 x^{2} + 90 x + 8

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

f (x) = 4 x^{3} - 24 x^{2} + 36 x

What is the extreme f(x)=x-1/2*y^2-1/3 x^3+y+6 ?
The extreme f(x)=x-1/2*y^2-1/3 x^3+y+6 is Saddle(-1,1),Maximum(1,1)