The function $f$ is defined for those $x$ for which the expressions under the even roots are non-negative, that is, $x \geq 0$, and $2x - 1/2 \geq 0$, which together means: for $x \geq 1/4$. Noting that $x_{1} x_{2} = 1$, and $21/4$, so $f$ is defined at both specified points.

$f(x_{1})$ is positive because all four factors are positive, while in the case of $f(x_{2})$, only the cube root is negative, the other three are positive, so $f(x_{2})$ is negative. After this, it is sufficient to calculate the absolute values of the two numbers, for which it is practical to first calculate the $f^6$ power, as this does not require taking a root. The two calculations can be combined because $x_{1}$ and $x_{2}$ differ only in the sign of $\sqrt{3}$, and the same applies to $(4-x)$, $(x-1)$, and $(2x-1/2)$ (in other words, $x_{1} + x_{2}$ is rational, and therefore the sum of the values of the other three factors at $x_{1}$ and $x_{2}$ is also rational). Now,

$$[f(2 \pm \sqrt{3})]^{6} = (2 \mp \sqrt{3})^{6} \cdot (2 \pm \sqrt{3})^{3} \cdot (1 \pm \sqrt{3})^{2} \cdot \frac{7 \pm 4 \sqrt{3}}{2}$$

and the right-hand side, by repeatedly applying the identity $(a+b)(a-b)=a^{2}-b^{2}$, squaring, and recognizing a perfect square in the last factor, transforms as follows:

$$(2 \mp \sqrt{3})^{3} \cdot [(2 \mp \sqrt{3}) \cdot (2 \pm \sqrt{3})]^{3} \cdot 2(2 \pm \sqrt{3}) \cdot \frac{(2 \pm \sqrt{3})^{2}}{2} = 1 = 1^{6}$$

since 1 stands in the square brackets, and the product of the remaining binomials is also 1. Therefore, $f(2+\sqrt{3}) = +1$ and $f(2-\sqrt{3}) = -1$.