The answer is no. 

Indeed, suppose that $n=1$ and $f(x)=e^{-x}$. Then 
$$X(t)=2\ln\cosh\frac{t}{\sqrt{2}}.$$

So, the velocity $X'(t)=\sqrt{2} \tanh \dfrac{t}{\sqrt{2}}$ is increasing with $t$, and/but
$X(t)\sim t\sqrt2\to\infty$ as $t\to\infty$. 

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Here is the graph $\{(t,X(t))\colon0\le t\le10\}$:

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/Z44v6.png