Verified answer

f' '(x) = -x / sqrt(1 - x^2)

critical points are where

function =0

function is undefined

function is undefined where the root is undefined

√(1-x^2) =0 when x=± 1

f(x) = sqrt(1 - x^2)

Re-write in exponent form:

f(x) = (1 - x^2)^(1/2)

Use the chain rule to find the derivative:

f '(x) = (1/2) * (-2x) * (1 - x^2)^(-1/2)

Re-write to simplify:

f '(x) = -x / sqrt(1 - x^2)

Set this equal to 0 to find critical points:

-x / sqrt(1 - x^2) = 0

The denominator gives us the spots where the function doesn't exist, meaning:

sqrt(1 - x^2) = 0, because a denominator cannot be 0. Solve by squaring both sides:

1 - x^2 = 0

Move the x^2 over:

x^2 = 1

Thus, x = -1 and x = 1.

Also, from the numerator, we simply have where the derivative is 0, and so:

-x = 0

So, x = 0 is your critical value.

Thus, the three critical values are x = 0, x = -1, and x = 1.

*Edited because I forgot to include the critical points where the function doesn't exist!*