Given the function f(y) = 1 − (3)/(y), find the domain of y for which the range of f(y) is in the interval (-2, 7).

$$1 - \frac{3}{y} > -2$$

Add 2 to both sides:

$$3 - \frac{3}{y} > 0$$

Multiply by $y$ (we must consider two cases for $y > 0$ and $y < 0$):
Case 1: $y > 0$

$$3y - 3 > 0$$ $$3y > 3$$ $$y > 1$$

So, for $y > 0$, the condition is $y > 1$.

Case 2: $y < 0$
When multiplying by a negative number, reverse the inequality sign:

$$3y - 3 < 0$$ $$3y < 3$$ $$y < 1$$

So, for $y < 0$, the condition is $y < 1$. Since we assumed $y < 0$, this means all $y < 0$ satisfy the inequality.

Combining both cases for $f(y) > -2$, the solution is $y < 0$ or $y > 1$.
In interval notation: $(-\infty, 0) \cup (1, \infty)$.

$$1 - \frac{3}{y} < 7$$

Subtract 1 from both sides:

$$-\frac{3}{y} < 6$$

Multiply by $-1$ and reverse the inequality sign:

$$\frac{3}{y} > -6$$

Multiply by $y$ (again, consider two cases):
Case 1: $y > 0$

$$3 > -6y$$ $$-\frac{3}{6} < y$$ $$y > -\frac{1}{2}$$

So, for $y > 0$, the condition is $y > -\frac{1}{2}$. Since we assumed $y > 0$, this means all $y > 0$ satisfy the inequality.

Case 2: $y < 0$
When multiplying by a negative number, reverse the inequality sign:

$$3 < -6y$$ $$-\frac{3}{6} > y$$ $$y < -\frac{1}{2}$$

So, for $y < 0$, the condition is $y < -\frac{1}{2}$.

Combining both cases for $f(y) < 7$, the solution is $y < -\frac{1}{2}$ or $y > 0$.
In interval notation: $(-\infty, -\frac{1}{2}) \cup (0, \infty)$.

We need $y$ to satisfy both conditions:
Condition 1: $y \in (-\infty, 0) \cup (1, \infty)$
Condition 2: $y \in (-\infty, -\frac{1}{2}) \cup (0, \infty)$

Let's find the intersection of these two sets:
The intersection of $(-\infty, 0)$ and $(-\infty, -\frac{1}{2})$ is $(-\infty, -\frac{1}{2})$.
The intersection of $(1, \infty)$ and $(0, \infty)$ is $(1, \infty)$.
The other intersections are empty.

Therefore, the domain for the given range is $y < -\frac{1}{2}$ or $y > 1$.
In interval notation, the domain is $\boxed{\left(-\infty, -\frac{1}{2}\right) \cup (1, \infty)}$.

An arrow diagram shows the mapping from the domain to the range.
Domain: $D = \left\{y \in \mathbb{R} \mid y < -\frac{1}{2} \text{ or } y > 1\right\}$
Range: $R = \{f(y) \in \mathbb{R} \mid -2 < f(y) < 7\}$

The arrow diagram can be represented as:

$$\text{Domain } \left(-\infty, -\frac{1}{2}\right) \cup (1, \infty) \quad \xrightarrow{f(y) = 1 - \frac{3}{y}} \quad \text{Range } (-2, 7)$$