f(x) = arcsinx- 2arctanx Find any relative extrema of the function

This function is defined on [-1, 1] and is continuously differentiable on (-1, 1). Its derivative is
f'(x) = 1/sqrt(1 - x^2) - 2/(1 + x^2).
Let's solve the equation  f'(x) = 0:
1/sqrt(1 - x^2) = 2/(1 + x^2), both sides are non-negative, hence it may be squared:
1/(1 - x^2) = 4/(1 + x^2)^2, which is equivalent to
1 + 2x^2 + (x^2)^2 = 4 - 4x^2,  or  (x^2)^2 + 6x^2 - 3 = 0.
This gives us x^2 = -3 +- sqrt(9 + 3), it must be non-negative so only "+" is suitable.
Thus x^2 = -3 + sqrt(12) = sqrt(3)(2 - sqrt(3)) which is lt1, and x_(1,2) = +-sqrt(sqrt(3)(2 - sqrt(3))) approx +-0.68.
Now consider the sign of f'(x). Near x=+-1 it tends to +oo and therefore is positive, at x=0 it is negative. Therefore f(x) increases from -1 to -sqrt(sqrt(3)(2 - sqrt(3))), decreases from -sqrt(sqrt(3)(2 - sqrt(3))) to +sqrt(sqrt(3)(2 - sqrt(3))) and increases again from +sqrt(sqrt(3)(2 - sqrt(3))) to 1.
The answer: f(-1) is a local one-sided minimum, f(1) is a local one-sided maximum, f(-sqrt(sqrt(3)(2 - sqrt(3)))) is a local maximum and f(sqrt(sqrt(3)(2 - sqrt(3)))) is a local minimum.