The function is
$$f(x) =\begin{cases} 1+e^x, & \text{if } x < 0 \\2+\arctan x, & \text{if } x \geq 0\end{cases}$$
Continuity:
On the interval $(-\infty, 0)$, $f(x)=1+e^x$ is an elementary function and thus continuous.
On the interval $[0, +\infty)$, $f(x) = 2 + \arctan x$ is also an elementary function and is continuous.
Thus the function is continuous on $(-\infty, x) \cup [0, +\infty)$, it remains to check if it is continuous at $x=0$. Considering the left and the right limits, we have
$$\lim\limits_{x \uparrow 0} f(x)=\lim\limits_{x \uparrow 0} 1 + e^x=2=\lim\limits_{x \downarrow 0} (2 + \arctan x)=\lim\limits_{x \downarrow 0} f(x)$$
and thus $f(x)$ is continous at $x=0$
Uniform continuity:
Here I'll use the following criteria:
If $f$ is continuous on $[a, +\infty)$ and $\lim\limits_{x \to +\infty} f(x) = L \in \mathbb R$ (i.e. L is a finite number), then $f$ is uniformly continuous on $[a, +\infty)$. (The analogous statement holds for $(-\infty, a]$).
$1+e^x$ is continous on $(-\infty, 0]$ (as an elementary function) and $\lim\limits_{x \to -\infty} 1 + e^x = 1$, and therefore by the above criteria it is also uniformly continuous on $(-\infty, 0]$ and all its subsets, in particular $(-\infty, 0)$.
$2+\arctan x$ is continuous on $[0, +\infty)$ (as an elementary function) and $\lim\limits_{x \to +\infty} (2 + \arctan x) = 2 + \frac{\pi}{2}$, so by the above criteria it is also uniformly continuous on $[0, +\infty)$.
We get that $f(x)$ is uniformly continous on its whole domain $(-\infty, 0] \cup [0, +\infty)$.
Is this reasoning alright? I'm curious if there's a better\right way to do it