Question

need help on the attachment

Answer 1

We need to know what f(x) is first

Answer 2

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

Answer 3

ok, critical numbers are when the f'(x)=0, so let's find f'(x):
wait, is it
\[f(x)=1-2x^2/\sqrt{1-x^2}\]or\[f(x)=(1-2x^2)/\sqrt{1-x^2}\]
????

Answer 4

your original problem I mean (please use parentheses)

Answer 5

its the first you wrote, without parentheses

Answer 6

oops! Actually it was asking to find all the critical points of f in (−1, 1).

Answer 7

all those points are in [-1,1], but if for the open interval (-1,1)
the set would be all those points EXCEPT -1 and 1, so
\[x=-\sqrt{2/3},0,\sqrt{2/3}\]

Answer 8

Oh hell, I dropped a minus sign and have to do it all over again. My bad... Let me do it again:

Answer 9

\[f(x)=1-2x^2(1-x^2)^{-1/2}\]
\[f'(x)=(-4x)(1-x^2)^{-1/2}+(-2x^2)(-1/2)(1-x^2)^{-3/2}(-2x)\]\[=(1-x^2)^{-3/2}[-4x(1-x^2)^2-2x^3]\]\[=(1-x^2)^{-3/2}[-4x(1-2x+x^2)-2x^3]\]\[=(1-x^2)^{-3/2}(-4x+8x^2-6x^3)\]\[=(1-x^2)^{-3/2}(-2x)(3x^2-4x+2)\]
the denominator shows critical points for
x={-1,1}
which are outside of the open interval (-1,1) and so do not count.

the (-2x) gives the critical point x=0 which is not a choice on your list for some reason (I don't think I made any mistakes)

and the quadratic solves to something complex, so something is wrong here... are you sure the problem is not

????????????