Linear Algebra Challenge Question involving Vector Spaces. Problem:
Let $V$ be the space of continuously differentiable maps $f:\mathbb{R}\to\mathbb{R}$ and let $W$ be the subspace of those maps $f$ for which $f(0)=f^\prime(0)= 0.$ Let $Z$ be the subspace of $V$ consisting of maps $x\to ax+b$, with $a,b \in \mathbb{R}$. Prove that $V=W\oplus Z.$
In order to solve the problem, we have to show that for all $v\in V$ $$v=w+z,$$ where $w\in W$ and $z\in Z.$ I don't know why this must be true, however I can show that $W\cap Z=\{0\}$ since if $z(x)=ax+b$ and $z\in W$ then $z(0)=z^\prime(0)=0$ implies that $a=b=0.$ and so the intersection of the two subspaces must be the zero function denoted by $0.$ 
How do I proceed to show that every function can be represented as the sum of the functions $w\in W$ and $z\in Z$? 

Hint Suppose there is a decomposition $V = W \oplus Z$, so that for any $v(x) \in C^1(\Bbb R)$ we have
$$v(x) = w(x) + z(x)$$
for $w(x) \in W, z(x) \in Z$.
Now, we want to express $w(x)$ and $z(x)$ in terms of $v(x)$, and the only information available is the definitions of $W, Z$. Since $W$ is characterized by the evaluation of functions (and their derivatives) at $x = 0$, this suggests evaluating our above expression at $x = 0$, giving
$$v(0) = w(0) + z(0) .$$
Now, what does the definition of $W$ tell us about $w(0)$?
