Construct a fifth-degree polynomial $p(x)$ in the interval ${\cal I} = [-3,3]$ that meets these conditions:
I am not sure if this polynomial exists.
On
If the roots of $p$ are $a<b<c<d<e$, then the subintervals of $[-3,3]$ where $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ is positive are $(a,b),(c,d),$ and $(e,3]$. To make the maximum value of $p(x)$ as small as possible, the idea is to make the intervals where it is positive very short. That means we want $a$ and $b$ to be very close together, $c$ and $d$ to be very close together, and $e$ to be very close to $3$.
So, fix $\epsilon>0$ and whatever values of $a$ and $c$ you want (with $a<c-\epsilon$ and $c+\epsilon<3-\epsilon$) and consider $p(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ where $b=a+\epsilon$, $d=c+\epsilon$, and $e=3-\epsilon$. Note that then if $x\in (a,b)$, $|x-a|$ and $|x-b|$ are at most $\epsilon$, so $|p(x)|<6^3\epsilon^2$ (the other factors in $p(x)$ have absolute value less than $6$, the diameter of $[-3,3]$). Similarly we have $|p(x)|<6^3\epsilon^2$ for all $x\in (c,d)$ and $|p(x)|<6^4\epsilon$ for all $x\in (e,3]$. So, as long as we pick $\epsilon$ small enough, $|p(x)|$ will always be at most $1/5$ on any interval where $p(x)$ is positive, and so $p(x)\leq 1/5$ on all of $[-3,3]$.
No such polynomial exists.
Let $p(x)$ be a monic polynomial of degree $5$. Let $q(t)=\dfrac{1}{3^5}p(3t)$.
Then $q(t)$ is a monic polynomial of degree $5$ and $$ \max_{x \in [-3,3]} |p(x)| = 3^5 \max_{t \in [-1,1]} |q(t)| \geq \dfrac{3^5}{2^{4}} > \frac15 $$ because the monic polynomial of degree $5$ with minimal norm in $[-1,1]$ is $\dfrac{1}{2^{4}}T_{5}(t)$, where $T_5$ is the $5$-th Chebyshev polynomial. The minimal norm is $\dfrac{1}{2^{4}}$.