Assume that $d$,$e$,$f$ are the lenghts of the sides of a triangle. Drawing sketches of a few different triangles shows that as the angle between $d$ and $e$ approaches 180 degree, $f$ approaches $d+e$ from below, and that as the angle between $d$ and $e$ approaches 0 degree, $f$ approaches $|d-e|$ from ablove. You thus get that $ |d-e| < f < d+e $ exactly if $d,e,f$ are the lenghts of the sides of a triangle.
Now lets look for intervals $I = (l, u) \subset \mathbb{R}$ where every combination $a,b,c \in I$ fullfills that inequality. Quite obviously, zero cannot be in $I$, since for $f=0$, there are no $d,e$ which fulfill the inequality. More generally, to ensure that $|d-e| < f$ always holds, you must have $l \geq u-l$, since $u-l$ is an upper bound for $|d-e|$ if $d,e \in I$. Similarly, $2l$ is a lower bound for $d+e$ if $d,e \in I$, and thus $u \leq 2l$ must hold if $f < d+e$ is to hold for all $d,e,f \in I$. Since $l \geq u-l$ is in fact equivalent to $2l \geq u$, you get that all $d,e,f \in [l,u]$ are the lenghts of the sides of some triangle iff $ \begin{eqnarray} u &\leq& 2l \end{eqnarray} $
Thus, you need to values of $m$ for which the range of $f$, i.e. the set $\{f(x):x\in\mathbb{R}\}$, is a subset of some inteval which satisfies the above. Can you do that?