Take a (convex) polyhedron: $$\mathcal{P} = \{ x \,|\, Ax \leq b \} $$ where $A \in \mathbb{R}^{m \times n},x \in \mathbb{R}^n, b \in \mathbb{R}^m$ and transform it using an element-wise monotonic function $f$ defined on $\mathbb{R}$, expressing the new set as: $$f(\mathcal{P}) = \{f(x) \,|\, Ax \leq b \} .$$
Can $f(\mathcal{P})$ be written as a polyhedron: $\{ x \vert Dx \leq c \} ,$ and how are $D$ and $c$ defined and calculated?
No. Let us consider the case $\mathcal{P} = \{ (t, 2\, t) \mid t \in [0,1]\}$ and $f(x) = \exp(x)$. Then, one can check that $\exp(\mathcal{P}) = \{ (x, x^2) \mid x \in [1,\mathrm{e}]\}$ which is not a polyhedron.