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

The extreme (sqrt(1-x^2))/(2x+1) is Maximum(-1,0),Minimum(1,0)

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

extreme points

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

Maximum (- 1, 0), Minimum (1, 0)

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

Plug

x = - 1

into

\frac{1 - x ^{2}}{2 x + 1} : 0

Maximum (- 1, 0)

Plug

x = 1

into

\frac{1 - x ^{2}}{2 x + 1} : 0

Minimum (1, 0)

Maximum (- 1, 0), Minimum (1, 0)

f (x) = \frac{4 x - 2}{x ^{2} - 10 x + 25}

f (x, y) = x y + 5 x - 5

f (x) = - x^{3} - 6 x^{2}

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

f (x) = x^{3} + 9 x^{2} + 24 x + 20

What is the extreme (sqrt(1-x^2))/(2x+1) ?
The extreme (sqrt(1-x^2))/(2x+1) is Maximum(-1,0),Minimum(1,0)