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

The critical (sqrt(1-x^2))/(2x+1) is x=-1,x=1

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

critical points

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

x = - 1, x = 1

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = - \frac{1}{2}, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{1 - x ^{2}}{2 x + 1} : - 1 \leq x < - \frac{1}{2} or - \frac{1}{2} < x \leq 1

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

is not defined at

x = - \frac{1}{2}

, therefore:

x = - 1, x = 1

f (x) = x + 7

f (x) = - x^{2} - 2 x + 5

f (x) = \frac{8}{3 + x}

f (x) = 3 + x

f (x) = (x - 2)^{2} - 1

What is the critical (sqrt(1-x^2))/(2x+1) ?
The critical (sqrt(1-x^2))/(2x+1) is x=-1,x=1