What is the extreme f(x,y)=(x^3)/3-xy^2+(2y^5)/5 ?

The extreme f(x,y)=(x^3)/3-xy^2+(2y^5)/5 is Minimum(1,1),Maximum(-1,-1)

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extreme f(x,y)=(x^3)/3-xy^2+(2y^5)/5

extreme points

f (x, y) = \frac{x ^{3}}{3} - x y^{2} + \frac{2 y ^{5}}{5}

Minimum (1, 1), Maximum (- 1, - 1)

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 8 y^{3} - 2 x

D (x, y) = 2 x (8 y^{3} - 2 x) - 4 y^{2}

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(- 1, - 1) :

Maximum

Minimum (1, 1), Maximum (- 1, - 1)

z = 20 x + 15 y

f (x) = - x^{2} + 7 x - 17

y = I n ∣ sec (5 x) + tan (5 x) ∣

f (x) = x^{4} - 5 x^{3}

f (x) = 1 + \frac{5}{x} - \frac{4}{x ^{2}}

What is the extreme f(x,y)=(x^3)/3-xy^2+(2y^5)/5 ?
The extreme f(x,y)=(x^3)/3-xy^2+(2y^5)/5 is Minimum(1,1),Maximum(-1,-1)