In a course on Intuitionistic Mathematics, the Intermediate Value Theorem is discussed, and it is shown why this theorem fails in intuitionism. This is true because:
Let $\rho$ be a real number for which we cannot decide $\rho<0\vee\rho=0\vee\rho>0.$ Then consider the function $f:[0,1]\rightarrow[0,1]$, for which $f(0)=0$, $f(1)=1$, $f(\frac13)=f(\frac23)=\frac12+\rho$ and linear inbetween. Then we cannot decide whether the $x$ for which $f(x)=\frac12$ satisfies $x>\frac13$ or $x<\frac23$.
Now, it is noted that the IVT does hold for something my lecturer calls ''locally variable'' (lokaal afwisselend in Dutch) functions, after which I wrote down the following:
Polynomial functions are locally variable, because:
For every $n\in\Bbb N$, for all real numbers $a_0,\ldots,a_n$ such that $a_n\neq 0$ (strongly), for all real numbers $b_0,\ldots,b_n$ that are pairwise (strongly) unequal, there is an $i\leq n$ such that $a_0+ a_1 b_i+\ldots+a_n b_i^n\neq 0$ (again strongly).
Could that be the definition of a local variable function, that there exists a real number for which the image is strongly unequal to 0? And then, how would one prove the IVT holds for these locally variable functions?