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

The monotone (sqrt(1-x^2))/x is Decreasing:-1<x<0,Decreasing:0<x<1

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

monotone intervals

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

Decreasing : - 1 < x < 0, Decreasing : 0 < x < 1

Solution steps

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

Find intervals:

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1

Decreasing : - 1 < x < 0, Decreasing : 0 < x < 1

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

f (x) = \frac{x}{x ^{2} + 64}

f (x) = x + 1 - x

(x + 2)^{\frac{1}{2}}

f (x) = \frac{( \frac{x}{x + 8} )}{( \frac{x}{x + 8} ) + 8}

What is the monotone (sqrt(1-x^2))/x ?
The monotone (sqrt(1-x^2))/x is Decreasing:-1<x<0,Decreasing:0<x<1