What is the extreme F(x)=(-(160x))/((x^2-16)^2) ?

The extreme F(x)=(-(160x))/((x^2-16)^2) is None

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extreme F(x)=(-(160x))/((x^2-16)^2)

extreme points

F (x) = \frac{- ( 160 x )}{( x ^{2} - 16 ) ^{2}}

None

Solution steps

f^{'} (x) = - \frac{160 ( - 3 x ^{2} - 16 )}{( x ^{2} - 16 ) ^{3}}

Find intervals:

None

None

x^{2} + y^{2} + 2 x - 4 y + 6

f (x) = - 6 x^{3} + 18 x^{2} + 54 x - 72

f (x) = - 15.4 x^{2} + 301.1 x

f (x) = - 3 x^{3} + 9 x^{2} + 10

f (x) = (x - y) (25 - x y)

What is the extreme F(x)=(-(160x))/((x^2-16)^2) ?
The extreme F(x)=(-(160x))/((x^2-16)^2) is None