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In chapter 7, equation 7.6 says CNOT works as follows:

CNOT: $frac1sqrt2 (|0rangle + |1rangle )otimes |xrangle rightarrow
frac1sqrt2 (|0rangle + (-1)^x |1rangle ) otimes |xrangle$, where it acts trivially if the target is $x=0$ state, and it flips the control if the target is the $x=1$ state.

I've looked at a few other resources about CNOT and this is the first time I encountered the $(-1)^x$ term.

Could someone explain to me where that term comes from?

Given that the matrix representation of CNOT is $$
beginpmatrix
1 & 0 &0 &0 \
0 & 1 & 0 & 0\
0 & 0 & 0 & 1\
0 & 0 & 1 & 0
endpmatrix$$ I don't see how that $(-1)^x$ came about.

Expanding on Jalex

Look at what happens on the possible terms.

begineqnarray*
mid 0 rangle otimes mid + rangle &to& mid 0 rangle otimes mid + rangle\
mid 0 rangle otimes mid - rangle &to& mid 0 rangle otimes mid - rangle\
mid 1 rangle otimes mid + rangle &to& mid 1 rangle otimes mid + rangle\
mid 1 rangle otimes mid - rangle &to& (-1) mid 1 rangle otimes mid - rangle\
endeqnarray*

where the first 2 are unchanged because the control is $0$ so nothing happens. The third is unchanged because NOT applied to $mid + rangle$ just gives back $mid + rangle$. The last is the only one with change because NOT applied to $mid - rangle$ gives $(-1) mid - rangle$.

We can summarize these possibilities by knowing that $mid + rangle$ goes with $x=0$ and $mid - rangle$ with $x=1$ as:

begineqnarray*
mid 0 rangle otimes mid x rangle &to& mid 0 rangle otimes mid x rangle\
mid 1 rangle otimes mid x rangle &to& (-1)^x mid 1 rangle otimes mid x rangle\
endeqnarray*

The first two become the first one above. And third and fourth, the second above.

Now add the two together along with a $frac1sqrt2$ prefactor to give

$$
frac1sqrt2 ( mid 0 rangle + mid 1 rangle ) otimes mid x rangle to frac1sqrt2 ( mid 0 rangle + (-1)^x mid 1 rangle ) otimes mid x rangle
$$

Here Preskill is using a physics convention that the states $|xrangle$ are the eigenstates of the $X$ operator. So $|xrangle$ with $x=0$ actually means $|+rangle$ and with $x=1$ actually means $|-rangle$.