Subspaces of $\mathbb{R}^{(0,3)}$ I'm working my way through Axler's "Linear Algebra Done Right", and I've run into a proposition that I don't understand.  Namely:

The set of differentiable real-valued functions $f$ on the interval (0, 3) such that $f'(2)=b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b=0$.

My intution is that this has something to do with the fact that a subspace requires an additive identity, but that because the domain of the differentiable functions is open and exludes 0, we have to make some concessions (i.e. $f'(2) = 0$), though I'm unsure of how to work up a proof that backs this.
Why must $f'(2) = 0$ to form a subspace of $\mathbb{R}^{(0,3)}$?
 A: Hint: 
$$(f+g)\in\text{That Set}\iff b=(f+g)'(2)=f'(2)+g'(2)=b+b.$$
A: I would just like to address the paragraph regarding intuition and the additive identity:  That intuition is incorrect.
First, we should define the object that we wish to verify is a subspace of $\mathbb{R}^{(0,3)}$ in something a little more precise than the natural language which it is first presented in.  We want to examine a set, $S$, where $$S=\{x,f\ |\ 0 \lt x \lt 3,\ f\ \epsilon\ \mathbb{R}^{(0,3)}\ \land\ f'(x) = 0\}$$
Specifically, we're charged with proving the restriction on the derivative of $f$, and this is best done in the manner alluded to by Workaholic.
The reason my initial intuition about the problem is ultimately incorrect has to do with the fact that $S$, as defined above, defines a functional space.  I was handicapping myself by thinking about it purely as a geometric space.  Consequently, I failed to see that the zero function (which is the additive identity in a functional space) is a natural member of $S$.
The reason $f'(x)$ must be zero for $f\ \epsilon\ S$ is that if it were otherwise $S$ would not be closed under addition, and therefore not a subspace of $\mathbb{R}^{(0,3)}$.
