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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

critical points

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

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

Solution steps

\frac{x ^{3}}{3} - x y^{2} + \frac{2 y ^{5}}{5}

Find where

\nabla f (x, y)

is equal to zero or undefined

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

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

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

f (x, y) = (x^{3} + 1) (y^{3} + 1)

x^{3} - 12 x^{2} + 36 x + 5

f (x) = x^{2} e^{20 x}

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