Number of Local Maxima Points for Logarithmic Function Transformation

Given $f(x) = \ln (g(x))$ for all $x \in \mathbf{R}$, where $g(x) \in (0, \infty)$, we can express $g(x)$ as:
$g(x) = e^{f(x)}$

Differentiating both sides with respect to $x$, we get:
$g'(x) = e^{f(x)} \cdot f'(x) = g(x) f'(x)$

Since $g(x) > 0$ for all $x \in \mathbf{R}$, the sign of $g'(x)$ is identical to the sign of $f'(x)$. Therefore, $g(x)$ has a local maximum at a point $x_0$ if and only if $f'(x)$ changes sign from positive to negative as $x$ increases through $x_0$.

We are given the first derivative of $f(x)$ as:
$f'(x) = 2010 (x - 2009)^1 (x - 2010)^2 (x - 2011)^3 (x - 2012)^4$

The critical points of $f(x)$ are $x = 2009, 2010, 2011, 2012$.

Let us analyze the sign of $f'(x)$ across the intervals formed by these critical points:

1. For $x < 2009$:

   • $(x - 2009) < 0$
   • $(x - 2010)^2 > 0$
   • $(x - 2011)^3 < 0$
   • $(x - 2012)^4 > 0$
   • Thus, $f'(x) > 0$.

2. For $2009 < x < 2010$:

   • $(x - 2009) > 0$
   • $(x - 2010)^2 > 0$
   • $(x - 2011)^3 < 0$
   • $(x - 2012)^4 > 0$
   • Thus, $f'(x) < 0$.

   Since $f'(x)$ changes sign from positive to negative as $x$ passes through $2009$, $f(x)$ (and consequently $g(x)$) has a local maximum at $x = 2009$.

3. For $2010 < x < 2011$:

   • $(x - 2009) > 0$
   • $(x - 2010)^2 > 0$
   • $(x - 2011)^3 < 0$
   • $(x - 2012)^4 > 0$
   • Thus, $f'(x) < 0$.

   Since $f'(x)$ does not change sign across $x = 2010$, $x = 2010$ is neither a local maximum nor a local minimum.

4. For $2011 < x < 2012$:

   • $(x - 2009) > 0$
   • $(x - 2010)^2 > 0$
   • $(x - 2011)^3 > 0$
   • $(x - 2012)^4 > 0$
   • Thus, $f'(x) > 0$.

   Since $f'(x)$ changes sign from negative to positive at $x = 2011$, $x = 2011$ is a point of local minimum.

5. For $x > 2012$:

   • All factors are positive, so $f'(x) > 0$.

   Since $f'(x)$ does not change sign across $x = 2012$, $x = 2012$ is neither a local maximum nor a local minimum.

From the above analysis, $g(x)$ has a local maximum at only one point, which is $x = 2009$.

Hence, the number of points in $\mathbf{R}$ at which $g$ has a local maximum is $1$.