What is the extreme f(x)=2x^3+2y^2-6x-8y+14 ?

The extreme f(x)=2x^3+2y^2-6x-8y+14 is Saddle(-1,2),Minimum(1,2)

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extreme f(x)=2x^3+2y^2-6x-8y+14

extreme points

f (x) = 2 x^{3} + 2 y^{2} - 6 x - 8 y + 14

Saddle (- 1, 2), Minimum (1, 2)

Solution steps

Find the critical points:

(- 1, 2), (1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 48 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(1, 2) :

Minimum

Saddle (- 1, 2), Minimum (1, 2)

f (x) = 2 x^{2} - 8 [0.6]

f (x) = x 16 - x^{2}, - 4 \leq x \leq 4

f (x) = \frac{4}{3} x^{3} + 3 x^{2} - 28 x - 44

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

f (x) = - x^{4} + 2 x^{3} + 8 x

What is the extreme f(x)=2x^3+2y^2-6x-8y+14 ?
The extreme f(x)=2x^3+2y^2-6x-8y+14 is Saddle(-1,2),Minimum(1,2)