Question: The following figure shows the devastating effect the opening of a new discount department store had on an established department store in a small town. The revenue of the discount store at time $t$ (in months) is given by $f(t)$ million dollars, whereas the revenue of the established department store at time $t$ is given by $g(t)$ million dollars. Answer the following questions by giving the value of $t$ at which the specified event took place.
The $t$ $y$-coordinate plane is given. Two curves are graphed.
The first curve labeled $y = f(t)$ begins at the origin, goes up and right becoming more steep, passes through $t = t_1$, passes through $t = t_2$, goes up and right becoming less steep, passes through $t = t_3$, and ends in the first quadrant nearly horizontal.
The second curve labeled $y = g(t)$ begins on the positive $y$-axis nearly horizontal, goes down and right becoming more steep, passes through $t = t_1$ crossing $f(t)$, passes through $t = t_2$ below $f(t)$, and ends on the $t$-axis at $t = t_3$.
(a)
The revenue of the established department store is decreasing at the slowest rate.
$t = $
1. **State the problem:** We need to find the value of $t$ at which the revenue of the established department store, represented by $g(t)$, is decreasing at the slowest rate.
2. **Understand the concept:** The rate of change of revenue is given by the derivative $g'(t)$. Since the revenue is decreasing, $g'(t) < 0$. The slowest rate of decrease means $g'(t)$ is negative but closest to zero (i.e., the maximum value of $g'(t)$ while still being negative).
3. **Analyze the graph description:** The curve $g(t)$ starts nearly horizontal on the positive $y$-axis, then goes down and right becoming more steep, passes through $t = t_1$ crossing $f(t)$, passes through $t = t_2$ below $f(t)$, and ends on the $t$-axis at $t = t_3$.
4. **Identify the slowest decreasing rate:** The slowest decrease corresponds to the point where the slope of $g(t)$ is least negative (closest to zero from below). From the description, $g(t)$ is nearly horizontal at the start (near $t=0$), then becomes steeper as $t$ increases.
5. **Conclusion:** The slowest rate of decrease occurs near the beginning of the graph, before $t_1$. Since the graph passes through $t_1$ crossing $f(t)$ and becomes steeper after, the slowest decrease is at $t = 0$ (or very close to it).
**Final answer:**
$$t = 0$$