What is the monotone (sqrt(1-x^2))/(2x+1) ?

The monotone (sqrt(1-x^2))/(2x+1) is Decreasing:-1<x<-1/2 ,Decreasing:-1/2 <x<1

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monotone (sqrt(1-x^2))/(2x+1)

monotone intervals

\frac{1 - x ^{2}}{2 x + 1}

Decreasing : - 1 < x < - \frac{1}{2}, Decreasing : - \frac{1}{2} < x < 1

Solution steps

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

Find intervals:

Decreasing

: - 1 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 1

Decreasing : - 1 < x < - \frac{1}{2}, Decreasing : - \frac{1}{2} < x < 1

x^{3} + 8

x = 1

f (x) = 3 x + 1

f (x) = \frac{x + 9}{x + 1}

\frac{1}{x ^{2} - 1}

What is the monotone (sqrt(1-x^2))/(2x+1) ?
The monotone (sqrt(1-x^2))/(2x+1) is Decreasing:-1<x<-1/2 ,Decreasing:-1/2 <x<1