To determine which root must also be a root of the polynomial function [tex]\( f(x) \)[/tex] with rational coefficients, given that the roots include 0, 4, and [tex]\( 3 + \sqrt{11} \)[/tex], we start by noting an important property of polynomials with rational coefficients.
Specifically, if a polynomial has rational coefficients and a root that involves a square root (that is, an irrational number), then its conjugate must also be a root to ensure that the coefficients remain rational. The conjugate of a number [tex]\( a + b\sqrt{c} \)[/tex] is [tex]\( a - b\sqrt{c} \)[/tex].
Given one of the roots is [tex]\( 3 + \sqrt{11} \)[/tex], its conjugate is [tex]\( 3 - \sqrt{11} \)[/tex]. Therefore, [tex]\( 3 - \sqrt{11} \)[/tex] must also be a root of the polynomial function [tex]\( f(x) \)[/tex].
Thus, the correct answer is:
[tex]\[ 3 - \sqrt{11} \][/tex]
