[tex]$f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = e^{x} + 1$[/tex]
[tex]\text{Range}(f) = \{y \in \mathbb{R} \vert x \in \mathbb{R}, f(x) = y\}[/tex]
[tex]\text{Suppose $f(x) = a$}\\e^{x} + 1 = a \implies e^{x} = a - 1 \implies x = \ln(a-1)\\[/tex]
[tex]\text{For the logarithm to be real, it must be true that $a - 1 > 0$}\\\text{But then we have that } a > 1. \implies \boxed{\text{Range}(f) = (1, \infty)}[/tex]
