Remark

When $I = E$, the image $f(I) = f(E)$ is the whole of $\mathbf{R}_+$, for if $b > \omega$, then $b^n$ is defined for all $n$, and therefore $n \cdot f(b)$ belongs to $f(E)$ for all $n$; this implies that $f(E)$ is not bounded above, because $f(b) > 0$.