`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...

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.