Let $E_N$ be the space of polynomials $P(x)$ of order $\leq N$. Prove there exist a constant $C_N>0$ so that $$\max\{|P(x)|,1\leq x\leq2\}\leq C_N·\max\{|P(x)|,0\leq x \leq 1\}, P\in E_N$$
What I've done: $P_1=\max\{|P(x)|,1\leq x\leq2\}=a_n2^n+a_{n-1}2^{n-1}+...+a_0$ where $a_n>0$ for all $n$. Then, $P_1\leq 2^n ·(a_n+a_{n-1}+a_{n-2}+...+a_0)=P_2=C_N·\max\{|P(x)|,0\leq x \leq 1\}$
Is this correct? If not, what is the correct answer? This is a question for students who just started a course in multivariable calculus. My answer does not use anything we have worked in class, that is why I am unsure of it.
It is certainly true that $$ \max_{1\le x\le2}|P(x)|\le2^N\sum_{k=0}^N|a_k|. $$ What is not so clear is an inequality in the other direction, that is $$ \max_{0\le x\le 1}|P(x)|\ge C_N\sum_{k=0}^N|a_k|. $$ If the coefficients are nonnegative, then it holds with $C_N=1$, but the general case needs some more work.
Since you are studying multivariable calculus, you have probably learned about norms, and that all norms are equivalent on finite dimensional vector spaces. Prove that for any interval $[a,b]$ $$ \|P\|_{a,b}=\max_{a\le x\le b}|P(x)| $$ is a norm on the $N+1$ dimensional vector space $E_N$ and you are done.