What is the extreme f(xy)=x^3+4y^2-3x+1 ?

The extreme f(xy)=x^3+4y^2-3x+1 is Saddle(-1,0),Minimum(1,0)

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extreme f(xy)=x^3+4y^2-3x+1

extreme points

f (x y) = x^{3} + 4 y^{2} - 3 x + 1

Saddle (- 1, 0), Minimum (1, 0)

Solution steps

Find the critical points:

(- 1, 0), (1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 48 x

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Check critical point

(1, 0) :

Minimum

Saddle (- 1, 0), Minimum (1, 0)

f (x) = \frac{x \frac{5}{3}}{2 + x}

f (x) = - 14 + 7 x - x^{2}

2 x 4 - x^{2}

f (x) = - x^{2} + 11, - 3 \leq x \leq 4

f (x) = sin (x) cos (x) + 2

What is the extreme f(xy)=x^3+4y^2-3x+1 ?
The extreme f(xy)=x^3+4y^2-3x+1 is Saddle(-1,0),Minimum(1,0)