How do you find the critical points(using partial derivative) for $f( x , y )=x^3 + y^2 - 6x^2 + y -1$ ?

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Answer 1 · 2015-08-20T10:27:08

Let Z = $x^3$ + $y^2$ - 6 $x^2$ + y -1

$f_x$ = 3 $x^2$ - 12x
$f_y$ = 2y + 1

$f_x$ = 0 $=>$ 3 $x^2$ - 12x = 0
x(3x - 12 ) = 0
x = 0
3x - 12 = 0
x = $12/3$ = 4

$f_y$ = 0 $=>$ 2y + 1 = 0

y = $-1/2$

How do you find the critical points(using partial derivative) for f( x , y )=x^3 + y^2 - 6x^2 + y -1f( x , y )=x^3 + y^2 - 6x^2 + y -1?