What is the extreme f(x)=(2^2)/(2^4+16) ?

The extreme f(x)=(2^2)/(2^4+16) is None

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extreme f(x)=(2^2)/(2^4+16)

extreme points

f (x) = \frac{2 ^{2}}{2 ^{4} + 16}

None

Solution steps

Find the critical points:

None

\frac{2 ^{2}}{2 ^{4} + 16}

has no critical points, therefore there are no extreme points

None

f (x, y) = 2 x^{2} - x y - 3 y^{2} - 3 x + 7 y

f (x, y) = x^{3} + 6 x y + y^{3}

f (x, y) = x y - 5 x + 15

f (x, y) = ln (4 x^{2} + 9 y^{2} + 36)

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

What is the extreme f(x)=(2^2)/(2^4+16) ?
The extreme f(x)=(2^2)/(2^4+16) is None