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

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

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

f (x) = x^{\frac{7}{9}}

x + cos (2 x), 0 \leq x \leq π

f (x) = \frac{x ^{2}}{x ^{2} + 192}

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

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