I'm working with the group of real linear functions ($f(x) = m·x + c, m \neq 0$) with the composition as the group product and via derivative I get that the generators of this group are:
$$\partial_mf = x, \quad \partial_cf = 1\tag1$$
From Eq. (1) we can write the Lie algebra as the set
$$\{r·x + k\ |\ r, k \in \mathbb{R}\}\tag2$$
which is not equal to the group because $r$ can be zero. But is this well done?
Thanks in advance!
Your Lie algebra as a set of maps to augment the identity map on your realization line is OK, but you should not infer that x is somehow significant. x is a prop of the realization on a line, and not a group parameter, so it is irrelevant to the generators: it is a placeholder of algebraic position, (m;c). In particular, this set of maps is not practical in specifying the unique Lie bracket of the algebra. See below.
The group is the simplest ever Affine group, in one dimension, a two-parameter nonabelian one; it is represented by upper-triangular 2×2 matrices with 1 in the second row, $$ \left( \begin{array}{c} f(x)\\ 1 \end{array}\right)= \left( \begin{array}{cc} m & c\\ 0 & 1 \end{array}\right) ~ \left( \begin{array}{c} x\\ 1 \end{array}\right). $$ Check the composition law $(m;c)\circ (n;d)=(mn;c+md)$.
Around the identity, (1;0), the respective generators are the gradients w.r.t. the parameters m-1, $$ A=\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right) , $$ and c, $$ B=\left( \begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right) $$ s.t. $~~~[A,B]=B $, the sparest prototype of a nonabelian Lie algebra.
You may then check that
$$ e^{aA+bB}= \left( \begin{array}{cc} e^a & \tfrac{b}{a}(e^a-1)\\ 0 & 1 \end{array}\right) , $$ that is, to say, $$ a=\ln m, \qquad b=\frac{c \ln m}{m-1} ~~. $$