\section*{Problem 5 - 051225}

Determine all real numbers \(x\) for which the polynomial

\[
f(x)=(x-1)(x-2)(x-3)(x-4)
\]

a) attains its smallest value (How large is this value?) and b) attains its largest value when \(x\) is restricted to the interval \(1 \leq x \leq 4\) (How large is this value?).
}

Shifting the function two units to the left results in:

\[
f(x)=(x+1.5) \cdot(x+0.5) \cdot(x-0.5) \cdot(x-1.5)=\left(x^{2}-2.25\right) \cdot\left(x^{2}-0.25\right)
\]

Setting \(f(x)=k\) yields:

\[
x^{4}-2.5 x^{2}+0.5625-k=0
\]

At local extrema, multiple solutions must occur.

\[
\left(x^{2}-1.25\right)^{2}=1+k
\]

Case 1: \(\left(x^{2}-1.25\right)^{2}=0: k=-1\) Minimum with multiple solutions,

Case 2: \(x^{2}=0: k=0.5625\) Maximum

In the interval [1,4], there are four roots, so there can be three intervals where local extrema can occur. Due to symmetry, the maximum 0.5625 can only occur once more or not at all. Therefore, there cannot be any larger values in the range [1,4], as there can only be three local extrema.