I know the range. Do I know the domain?

We will use an auxiliary function $g:\mathbb{R}\to\mathbb{R}$ , defined by the same formula as $f$ , i.e. $g(x)=2018^x+x^{2019}$ , for all $x\in \mathbb{R}$ . Notice that this new function is defined over the whole set of real numbers.

Now, taking the derivative of $g$ , we have:

$g^{\prime}(x)=2018^x\ln2018+2019x^{2018}>0$ , for all $x\in \mathbb{R}$

$\Rightarrow\, g$ is increasing

$\Rightarrow\, g$ is injective (one-to-one).

This means that if $a\not=b$ then $g(a)\not=g(b)$ .

Assume $A\not=\mathbb{R}$ . Then, there exists $x_0\in \mathbb{R}:\, x_0\not\in A$

Thus, for all $x\in A$ we have:

$x\not=x_0\Rightarrow\,g(x)\not=g(x_0)\Rightarrow\, f(x)\not=g(x_0)\Rightarrow\, g(x_0)\not\in f(A)$ .

But the latter is not true, since $f(A)=\mathbb{R}$ .

Therefor we conclude that $\color{#D61F06}{\boxed{A=\mathbb{R}}}$ , which furthermore means that finally $f$ is a bijection.