Consider the ring $R=\{f:\mathbb{R}\to\mathbb{R}\thinspace|\thinspace f\text{ differentiable}\}$, and let $I$ be the ideal defined by $I=\{f\in R\thinspace |\thinspace f(2)=f'(2)=0\}$. Find a map $R\to\mathbb{R}[x]/(x^2)$ to show that $R/I$ is isomorphic to $\mathbb{R}[x]/(x^2)$.
Since $f$ is allowed to be an arbitrary differentiable function, I cannot think of a natural map off the top of my head. I know I am going to eventually show the kernel of my map is $I$ and apply the First Isomorphism Theorem. Can anyone give me a hint on a map to pick?
Hint: Taylor polynomial.
Consider $f\mapsto f(2)+f'(2)(x-2)$.