Basic question on bra-ket notation Which of the following corresponds to a $ \psi(x)$, a wavefunction written in the position basis: $ x| \psi\rangle  $ or  $ \langle x| \psi\rangle  $? If it is the second expression (which my textbook asserts), what is the meaning of the first expression? 
 A: It is the second, $\psi(x) =  \langle x|\psi\rangle$ which is correct. The first, if $x$ is the position operator, is just the position operator acting on the state $|\psi\rangle$. The abstract state $|\psi\rangle$ can be expanded in any basis, using a completion relation: $$|\psi\rangle = \underbrace{\sum_i |i\rangle \langle i|}_{1~=~identity}\psi\rangle$$
(where the sum can mean sum or integral, depending on this situation) using some complete basis $\{|i\rangle\}$. An example of this is the position basis, $\{|x\rangle\}$, from which we have
$$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$
This shows us that the wavefunction $\psi(x) = \langle x|\psi\rangle$, is the coefficient from the expansion of the state $|\psi\rangle$ in the position basis $|x\rangle$, at position $x$.

Edit to respond to a question asked by Noah in response to Matt's answer:
"But doesn't an operator acting on the state project the state onto the eigenstates of the operator and is that the same as representing the state in a new basis?"
As I have shown above, the "projection" you are talking about is given by applying the identity in terms of the position basis $\mathbf{1=\int dx~|x\rangle\langle x|}$.
Let's see what happens when we apply $\hat{x}$ instead: Just like state vectors can be expanded in a complete basis, so too can operators. In general, if we take our complete basis $\{|i\rangle\}$ we can write the operator $\hat{A}$ in terms of matrix elements
$$\hat{A} = \sum_{i,j} |i\rangle\langle i |\hat{A}|j\rangle\langle j |$$
where the matrix elements are $A_{ij} = \langle i |\hat{A}|j\rangle$. For the case of the position operator $\hat{x}$ and using the position basis $\{|x\rangle\}$
$$\hat{x} = \int dx~dy~ |x\rangle\langle x |\hat{x}|y\rangle\langle y |\\
= \int dx~ x~|x\rangle\langle x |$$
that is, it is diagonal in the position basis (obviously). Using this we see
$$\hat{x}|\psi\rangle = \int dx~ x~|x\rangle\langle x |\psi\rangle$$
So we see that application of $\hat{x}$ can be thought of as a kind of projection, weighted by $x$. So we see that it is not the same as representing the state in the $x$ basis, which is actually
$$|\psi\rangle = \int dx~ |x\rangle \langle x|\psi\rangle$$
as above.
A: $\psi(x)\equiv\left\langle{x}\,\middle|\,\psi\right\rangle$ is the correct notation.  $x\left|\psi\right\rangle$ means that the position operator is acting on the state.  People sometimes put a hat on operators to remove ambiguity:  $\hat{x}\left|\psi\right\rangle$.
