Answer :
Let's analyze the given piecewise function [tex]\(D(t)\)[/tex] that defines the traveler's distance from home, in miles, as a function of time, in hours.
The function is defined as:
[tex]\[ D(t) = \begin{cases} 300t + 125, & \text{if } 0 \leq t < 2.5 \\ 875, & \text{if } 2.5 \leq t \leq 3.5 \\ 75t + 612.5, & \text{if } 3.5 < t \leq 6 \end{cases} \][/tex]
Now, let's verify the distances at specific times using the given function:
1. Starting distance at 0 hours:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
Therefore, the starting distance at 0 hours is 125 miles, not 300 miles. This means the option "The starting distance, at 0 hours, is 300 miles" is incorrect.
2. Distance at 2 hours:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
Therefore, at 2 hours, the traveler is 725 miles from home, making the option "At 2 hours, the traveler is 725 miles from home" correct.
3. Distance at 2.5 hours:
According to the function definition, for [tex]\(2.5 \leq t \leq 3.5\)[/tex]
[tex]\[ D(t) = 875 \text{ miles} \][/tex]
Therefore, at 2.5 hours, the distance is constant at 875 miles. The statement "At 2.5 hours, the traveler is still moving farther from home" is incorrect because the distance is constant, not increasing.
4. Distance at 3 hours:
Using the same interval as above, we get:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
Hence, at 3 hours, the distance is constant at 875 miles, so the option "At 3 hours, the distance is constant, at 875 miles" is correct.
5. Total distance from home after 6 hours:
According to the function definition, for [tex]\(3.5 < t \leq 6\)[/tex]
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
Therefore, the total distance from home after 6 hours is 1062.5 miles, making the statement "The total distance from home after 6 hours is [tex]$1,062.5$[/tex] miles" correct.
In summary, the three correct options are:
1. At 2 hours, the traveler is 725 miles from home.
2. At 3 hours, the distance is constant, at 875 miles.
3. The total distance from home after 6 hours is 1,062.5 miles.
The function is defined as:
[tex]\[ D(t) = \begin{cases} 300t + 125, & \text{if } 0 \leq t < 2.5 \\ 875, & \text{if } 2.5 \leq t \leq 3.5 \\ 75t + 612.5, & \text{if } 3.5 < t \leq 6 \end{cases} \][/tex]
Now, let's verify the distances at specific times using the given function:
1. Starting distance at 0 hours:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
Therefore, the starting distance at 0 hours is 125 miles, not 300 miles. This means the option "The starting distance, at 0 hours, is 300 miles" is incorrect.
2. Distance at 2 hours:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
Therefore, at 2 hours, the traveler is 725 miles from home, making the option "At 2 hours, the traveler is 725 miles from home" correct.
3. Distance at 2.5 hours:
According to the function definition, for [tex]\(2.5 \leq t \leq 3.5\)[/tex]
[tex]\[ D(t) = 875 \text{ miles} \][/tex]
Therefore, at 2.5 hours, the distance is constant at 875 miles. The statement "At 2.5 hours, the traveler is still moving farther from home" is incorrect because the distance is constant, not increasing.
4. Distance at 3 hours:
Using the same interval as above, we get:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
Hence, at 3 hours, the distance is constant at 875 miles, so the option "At 3 hours, the distance is constant, at 875 miles" is correct.
5. Total distance from home after 6 hours:
According to the function definition, for [tex]\(3.5 < t \leq 6\)[/tex]
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
Therefore, the total distance from home after 6 hours is 1062.5 miles, making the statement "The total distance from home after 6 hours is [tex]$1,062.5$[/tex] miles" correct.
In summary, the three correct options are:
1. At 2 hours, the traveler is 725 miles from home.
2. At 3 hours, the distance is constant, at 875 miles.
3. The total distance from home after 6 hours is 1,062.5 miles.