Is $R[X]/(f)$ Cohen-Macaulay if $R$ is so?

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Let $R$ be a commutative (Noetherian) Cohen-Macaulay ring, and $f \in R[X]$ be monic. I guess that $R[X]/(f)$ is also Cohen-Macaulay. Is my hunch valid? Thanks for any help.

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Proposition: If $S$ is CM, and $f \in S$ is a nonzerodivisor, then $S/(f)$ is CM.

Proof: Choosing a maximal ideal of $S/(f)$, we may assume $(S, m)$ is local. Then $\text{depth}(S/(f)) = \text{grade}(m/(f)) \ge \text{grade}(m) - 1 = \text{depth}(S) - 1$, while $\dim S/(f) \le \dim S - 1$.

Now, if $R$ is CM, then $R[x]$ is CM, and if $f$ is monic, then $f$ is a nonzerodivisor in $R[x]$.