= Solution
Now $B=\overline{\mathbb C[1,x]}$. Let $U$ be a bounded component of $\rho_A(x)$ and $\lambda\in U$. Its <boundary> lies in $\sigma_A(x)$: an <open> disk about a <boundary> point in the <algebra resolvent set> would meet $U$ and would belong to the same component, contradicting <boundary> membership. The <maximum modulus principle> on the bounded domain $U$ therefore gives, for every complex <polynomial> $p$,
$$
|p(\lambda)|\le\max_{\partial U}|p|
\le\max_{\sigma_A(x)}|p|=r_A(p(x))\le\|p(x)\|.
$$
Here <spectral mapping theorem> follows by factoring $p(z)-\mu$: its commuting factors are invertible exactly when every root avoids the <Banach algebra spectrum>.
The rule $p(x)\mapsto p(\lambda)$ is well-defined by the displayed bound, multiplicative and bounded on the dense <polynomial> <subalgebra>. It extends to a <unital> <character> $\varphi_\lambda$ on $B$. Since $\varphi_\lambda(\lambda1-x)=0$, that element cannot have an inverse in $B$. Thus \b[$U\subseteq\sigma_B(x)$]. Together with part (iii), this proves that the <Banach algebra spectrum> of a polynomial-generated Banach <subalgebra> is exactly the original <Banach algebra spectrum> with all its bounded complementary components filled in.
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