What is the extreme f(x)=(0.9)/(1+6e^{-0.32x)} ?

The extreme f(x)=(0.9)/(1+6e^{-0.32x)} is None

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extreme f(x)=(0.9)/(1+6e^{-0.32x)}

extreme points

f (x) = \frac{0.9}{1 + 6 e ^{- 0.32 x}}

None

Solution steps

f^{'} (x) = \frac{1.728 e ^{- 0.32 x}}{( 1 + 6 e ^{- 0.32 x} ) ^{2}}

Find intervals:

None

None

f (x, y) = x^{2} + 3 x y + 4.5 y^{2} + 4 x

5 x^{3} - 3 x^{5}

S (u, t) = \frac{u + u ^{0}}{2} \cdot t

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

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

What is the extreme f(x)=(0.9)/(1+6e^{-0.32x)} ?
The extreme f(x)=(0.9)/(1+6e^{-0.32x)} is None