What is the extreme y=(-8-8x^2)/(2sqrt(x)(1+x)) ?

The extreme y=(-8-8x^2)/(2sqrt(x)(1+x)) is Maximum(1,-4)

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extreme y=(-8-8x^2)/(2sqrt(x)(1+x))

extreme points

y = \frac{- 8 - 8 x ^{2}}{2 x ( 1 + x )}

Maximum (1, - 4)

Solution steps

f^{'} (x) = - \frac{2 ( x ^{3} + 3 x ^{2} - 3 x - 1 )}{x x ( x + 1 ) ^{2}}

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

\frac{- 8 - 8 x ^{2}}{2 x ( 1 + x )} : - 4

Maximum (1, - 4)

f (t) = \frac{e ^{t}}{t ^{2} + 1}

\frac{x + 3}{x - 2}

4 cos (3 x) + 5

f (x) = \frac{e ^{x}}{x ^{8}}

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

What is the extreme y=(-8-8x^2)/(2sqrt(x)(1+x)) ?
The extreme y=(-8-8x^2)/(2sqrt(x)(1+x)) is Maximum(1,-4)