The function [tex]\( D(t) \)[/tex] defines a traveler's distance from home, in miles, as a function of time, in hours.

[tex]\[ D(t) = \begin{cases}
300t + 125, & 0 \leq t \ \textless \ 2.5 \\
875, & 2.5 \leq t \leq 3.5 \\
75t + 612.5, & 3.5 \ \textless \ t \leq 6
\end{cases} \][/tex]

Which times and distances are represented by the function? Select three options.

A. The starting distance, at 0 hours, is 300 miles.
B. At 2 hours, the traveler is 725 miles from home.
C. At 2.5 hours, the traveler is still moving farther from home.
D. At 3 hours, the distance is constant, at 875 miles.
E. The total distance from home after 6 hours is 1,062.5 miles.



Answer :

To solve this question, let's evaluate the traveler's distance from home using the given piecewise function [tex]\( D(t) \)[/tex] at specific time points.

1. The starting distance, at 0 hours:

For [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300 \cdot 0 + 125 = 125 \][/tex]

Therefore, the starting distance is 125 miles, not 300 miles. This means the statement "The starting distance, at 0 hours, is 300 miles" is false.

2. At 2 hours, the traveler is 725 miles from home:

For [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300 \cdot 2 + 125 = 600 + 125 = 725 \][/tex]

Therefore, the statement "At 2 hours, the traveler is 725 miles from home" is true.

3. At 2.5 hours, the traveler is still moving farther from home:

Just before [tex]\( t = 2.5 \)[/tex], we need to check if the function value is increasing:
- For [tex]\( t \)[/tex] values just less than 2.5, use [tex]\( D(t) = 300t + 125 \)[/tex].

As [tex]\( t \)[/tex] approaches 2.5 from the left:
[tex]\[ D(2.5 - \epsilon) = 300(2.5 - \epsilon) + 125 \][/tex]
where [tex]\( \epsilon \)[/tex] is a very small positive number.

Since [tex]\( D(t) \)[/tex] is continuous and increasing in the interval [tex]\([0, 2.5)\)[/tex], as [tex]\( t \)[/tex] approaches 2.5, [tex]\( D(t) \)[/tex] gets closer to 875.

Therefore, the statement "At 2.5 hours, the traveler is still moving farther from home" is true.

4. At 3 hours, the distance is constant, at 875 miles:

For [tex]\( t = 3 \)[/tex]:
[tex]\[ D(3) = 875 \][/tex]

Therefore, the statement "At 3 hours, the distance is constant, at 875 miles" is true.

5. The total distance from home after 6 hours is 1,062.5 miles:

For [tex]\( t = 6 \)[/tex]:
[tex]\[ D(6) = 75 \cdot 6 + 612.5 = 450 + 612.5 = 1062.5 \][/tex]

Therefore, the statement "The total distance from home after 6 hours is [tex]$1,062.5$[/tex] miles" is true.

Given these evaluations, the correct statements are:

1. At 2 hours, the traveler is 725 miles from home.
2. At 2.5 hours, the traveler is still moving farther from home.
3. At 3 hours, the distance is constant, at 875 miles.
4. The total distance from home after 6 hours is [tex]$1,062.5$[/tex] miles.

The three correct options from the list are:

- At 2 hours, the traveler is 725 miles from home.
- At 2.5 hours, the traveler is still moving farther from home.
- At 3 hours, the distance is constant, at 875 miles.

Another valid option is the total distance after 6 hours being 1,062.5 miles although the question asked for three options only.

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