You need to use the first derivative test to evaluate the critical points of the given function, such that:

`f'(x) = ((x^4-x^2+2)'(x^4+2x^2+1) - (x^4-x^2+2)(x^4+2x^2+1)')/((x^4+2x^2+1)^2)`

`f'(x) = ((4x^3 - 2x)(x^4+2x^2+1) - (4x^3 + 4x)(x^4-x^2+2))/((x^4+2x^2+1)^2)`

`f'(x) = (4x^7 + 8x^5 + 4x^3 - 2x^5 - 4x^3 - 2x - 4x^7 + 4x^5 - 8x^3 - 4x^5 + 4x^3 - 8x)/((x^4+2x^2+1)^2)`

Reducing duplicate terms yields:

`f'(x) = (6x^5 - 4x^3 - 10x)/((x^4+2x^2+1)^2)`

You need to set the equation equal to zero, such that:

`(6x^5 - 4x^3 - 10x)/((x^4+2x^2+1)^2) = 0 => {(6x^5 - 4x^3 - 10x = 0),((x^4+2x^2+1)^2 != 0):}`

Solving the top equation yields:

`6x^5 - 4x^3 - 10x = 0 => 2x(3x^4 - 2x^2 - 5) = 0`

sing zero product rule, yields:

`2x = 0 => x = 0`

`3x^4 - 2x^2 - 5 = 0`

You should come up with the following substitution that helps you to evaluate the quartic equation above, such that:

`x^2 = t => 3t^2 - 2t - 5 = 0`

Using quadratic formula, yields:

`t_(1,2) = (2+-sqrt(4 + 60))/6 => t_(1,2) = (2+-8)/6`

`t_1 = 5/3; t_2 = -1`

Replacing back `x^2` for t yields:

`x^2 = 5/3 => x_(1,2) = +-sqrt(5/3)`

`x^2 = -1` invalid for `AA` `x in R`

Hence, evaluating the sign of the expression `f'(x)` yields that is negative over `(-oo,-sqrt(5/3))U(0,sqrt(5/3))` and positive over `(-sqrt(5/3),0)U(sqrt(5/3),oo)` .

**Hence, the function reaches its minimum point twice, at `x = -sqrt(5/3)` and **`x = sqrt(5/3).`