Show that $\sqrt{x^2+x+2}$ is defined and continuous 
Show that the function $g(x)=\sqrt{x^2+x+2}$ 
  is defined and is continuous on $\mathbb{R}$.

I have tried completion of square for $$x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}$$
This means that range, $r\ge 7/4$ in domain $\mathbb{R}$.
Cannot find any more logic in it, please help.
 A: You need the expression within the square root is non-negative, and you already showed this since $\;x^2+x+2=\left(x+\frac12\right)^2+\frac74\;\leftarrow$ this is the sum of two non-negative expressions and thus is always non-negative, so your function's defined in all of $\;\Bbb R\;$.
And it is continuous in all of $\;\Bbb R\;$ because it is the composition of two continuous functions: square root and a polynomial.
A: You have already shown that the inner function $x^2+x+2\geqslant 7/4$. The function $\sqrt t$ is defined for $t\geqslant 0$ and continuous for $t>0$. If you have
$$t=x^2+x+2$$ then the function $\sqrt t$ is defined and continuous as $t$ is continuous and $t\geqslant 7/4$ (here $t$ is a function of $x$).
A: To prove that $g (x)$ is defined on $\mathbf {R}$, we need to prove that for any $x_0 \in \mathbf {R}$ the value $g (x_0)$ exists. Indeed, since $x^2 + x + 2 > 0$ for any $x \in \mathbf {R}$, the expression $$g (x) = \sqrt {x^2 + x + 2}$$ is meaningful. Additionally, $g (x)$ is always $> \sqrt {7}/2$.
We now prove the continuity. Let $c$ be an arbitrary real number, and consider the limit $$\lim_{x \to c} g (x) = \lim_{x \to c} \sqrt {x^2 + x + 2} = \sqrt {c^2 + c + 2} = g (c).$$ Since $g$ is defined on reals, $g (c)$ exists. The limit also exists, since left and right limits exist and are equal. Also, the value of the limit equals $g (c)$. These three prove that $g (x)$ is continuous at every $c \in \mathbf {R}$, that is, it is continuous on $\mathbf {R}$.
