Step 1: Rewrite the function in a useful form.
We are asked for the range of $f(x) = \dfrac{x^2+x+1}{x}$ where $x \neq 0$. Divide each term in the numerator by $x$: \[ f(x) = x + 1 + \frac{1}{x} \] Let $y = x + 1 + \frac{1}{x}$, so $y - 1 = x + \frac{1}{x}$.
Step 2: What values can $x + \frac{1}{x}$ take?
This is a classic result. Think of it this way: if $x > 0$, by AM-GM, $x + \frac{1}{x} \geq 2\sqrt{x \cdot \frac{1}{x}} = 2$. So the minimum is $2$ (achieved at $x=1$). If $x < 0$, let $x = -t$ where $t > 0$. Then $x + \frac{1}{x} = -t - \frac{1}{t} \leq -2$. So the maximum is $-2$ (at $x = -1$).
Step 3: Convert back to values of $y$.
We have $y - 1 = x + \frac{1}{x}$. For $x > 0$: \[ y - 1 \geq 2 \implies y \geq 3 \] For $x < 0$: \[ y - 1 \leq -2 \implies y \leq -1 \]
Step 4: Verify the boundary values are achieved.
At $x = 1$: $f(1) = 1 + 1 + 1 = 3$. Boundary $y=3$ is included. At $x = -1$: $f(-1) = -1 + 1 - 1 = -1$. Boundary $y = -1$ is included. Both ends are closed, so brackets are correct.
Step 5: Values between $-1$ and $3$ are excluded.
Since for $x > 0$, $y \geq 3$ always, and for $x < 0$, $y \leq -1$ always, there is no $x$ that gives $-1 < y < 3$. That gap is completely missing from the range.
Step 6: Write the final range.
Combining both parts: \[ \boxed{(-\infty,-1]\cup[3,\infty)} \]