What is the extreme f(x)=3x^2-3y^2+6=0 ?

The extreme f(x)=3x^2-3y^2+6=0 is None

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extreme f(x)=3x^2-3y^2+6=0

extreme points

f (x) = 3 x^{2} - 3 y^{2} + 6 = 0

None

Solution steps

Find the critical points:

None

0

has no critical points, therefore there are no extreme points

None

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

f (x, y) = 2 x^{2} + y^{2} - 8 x + 8 y

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

x^{4} - 8 x^{3} + 7 x^{2} + 6 x - 4

f (x, y) = 5 - (x - 6)^{2} - (y + 2)^{2}