What is the extreme f(x)=x^2+2y^2x^2+y^2=1 ?

The extreme f(x)=x^2+2y^2x^2+y^2=1 is None

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extreme f(x)=x^2+2y^2x^2+y^2=1

extreme points

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

None

Solution steps

Find the critical points:

None

1

has no critical points, therefore there are no extreme points

None

f (x) = \frac{4 x + 2}{3 x - 6}

(\frac{1}{2})^{x - 1} + 5

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

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

What is the extreme f(x)=x^2+2y^2x^2+y^2=1 ?
The extreme f(x)=x^2+2y^2x^2+y^2=1 is None