At least there is a simple result for the case $f(x)=e^{tx}$. Under suitable conditions, we have
$$
{\rm corr}(X,e^{tX} ) = \frac{{{\rm E}[Xe^{tX} ] - \mu _X {\rm E}[e^{tX} ]}}{{\sigma _X \sqrt {{\rm E}[e^{2tX} ] - {\rm E}^2 [e^{tX} ]} }} = \frac{{m_X '(t) - \mu _X m_X (t)}}{{\sigma _X \sqrt {m_X (2t) - m_X^2 (t)} }},
$$
where $m_X (\cdot)$, $\mu_X$, and $\sigma^2_X$ denote the moment-generating function, the expectation, and the variance of $X$, respectively.

For example, for $X$ exponential with mean $1/\lambda$ (hence $m_X (\tau) = \frac{\lambda }{{\lambda  - \tau}}$, $\tau \lt \lambda$, $\sigma_X = 1/\lambda$), this yields
$$
{\rm corr}(X,e^{tX} ) = \frac{{\lambda t}}{{(\lambda  - t)^2 \sqrt {\frac{\lambda }{{\lambda  - 2t}} - (\frac{\lambda }{{\lambda  - t}})^2 } }}
, \;\; t \in ( - \infty ,\lambda /2)\backslash \lbrace 0 \rbrace .
$$
(Using ${\rm E}[X^n] = n!/\lambda^n$, we can also find that ${\rm corr}(X,X^n) = \frac{n}{{\sqrt {{2n \choose n} - 1} }}$.)

Interestingly, $\lim _{t \to 0 \pm  } {\rm corr}(X,e^{tX} ) =  \pm 1$ and $\lim _{t \uparrow \lambda /2} {\rm corr}(X,e^{tX} ) = 0$. Can you explain these results?

The general formula for ${\rm corr}(X,e^{tX} )$ may help you reach some conclusion. In particular, note that the functions $f(x)=e^{tx}$ and $m_X (t)$ are often closely related, see [here][1], for example. 

  [1]: http://en.wikipedia.org/wiki/Moment-generating_function#Examples