Question

a remote - control car moves toward a wall, stops, backs up, and stops. the graph shows the cars distance from the wall as a function of time. answers parts a through d.
a. write the domain and range in inequalities. identify the domain. select the correct choice below and fill in the answer box(es) to complete your choice.
a
b. ( xleq)
c. ( leq x)
d. ( yleq)
e. ( 0leq xleq10)
f. ( leq y)
identify the range. select the correct choice below and fill in the answer box(es) to complete your choice.
a. ( leq x)
b. ( leq xleq)
c. ( leq y)
d. ( xleq)
e. ( leq yleq)
f. ( yleq)

Answer:

To solve the problem about the remote - control car's distance from the wall as a function of time, we analyze the domain and range:

Part a: Domain

The domain of a function represents the set of all possible input values (in this case, time \(x\)). From the context of the car's motion (starting, moving, stopping, reversing, and stopping again) and the graph (even though we can infer from the typical motion of such a scenario), the time starts at \(0\) and goes up to a maximum value (let's assume from the inequality option \(0\leq x\leq10\) that the total time of the motion is \(10\) seconds). So the domain is the set of all \(x\) such that \(0\leq x\leq10\), which corresponds to option E.

Part b: Range

The range of a function represents the set of all possible output values (in this case, distance from the wall \(y\)). When the car moves, the distance from the wall will have a minimum and a maximum value. Let's assume that the minimum distance is \(0\) (when the car is closest to the wall) and the maximum distance is some value (let's say \(y_{max}\)). The range will be the set of all \(y\) values between the minimum and maximum distance. If we consider the general form of the range for such a motion, it will be of the form \(a\leq y\leq b\) (where \(a\) is the minimum distance and \(b\) is the maximum distance). Looking at the options, if we assume the minimum distance is \(0\) and the maximum distance is, for example, \(y_{max}\), the range inequality will be of the form \(0\leq y\leq y_{max}\) (or similar). But from the given options, if we take the typical motion, the range is described by an inequality of the form \(s\leq y\leq t\) (where \(s\) and \(t\) are the minimum and maximum distances). If we assume that the minimum distance is \(0\) and the maximum distance is, say, \(10\) (matching the domain's upper limit in a consistent problem - setting), and looking at the options, the range would be represented by an inequality like \(0\leq y\leq10\) (but we have to match the option format). If we consider the options for range:

• Option A: \(s\leq y\leq t\) (but we need to fill in \(s\) and \(t\))
• Option C: \(s\leq y\) (this is not a closed range, so incorrect as the car stops at two points, so distance has a minimum and maximum)
• Option E: \(0\leq x\leq10\) (this is domain, not range)
• Option B: \(s\leq x\leq t\) (this is domain, not range)
• Option D: \(y\leq s\) (incorrect, as distance can't be only less than or equal to a value without a lower bound in this motion)
• Option F: \(s\leq y\) (incorrect, same as option C)

If we assume that the minimum distance from the wall is \(0\) (when the car is closest) and the maximum distance is \(10\) (to be consistent with the domain's \(0 - 10\) time range in a simple problem - setting), and the range is \(0\leq y\leq10\), but from the given options, if we have to choose the format, and assuming that the range is of the form \(a\leq y\leq b\), and if we take the option that represents a closed interval for \(y\), and if we assume the minimum distance is \(0\) and maximum is \(10\), the range inequality is \(0\leq y\leq10\) (but we have to match the option. If we consider the options, and assume that the range is \(0\leq y\leq10\), and the option for range with the correct format (if we take the first part of the range options) is option A (but we need to fill in the blanks). However, if we look at the problem again, since the domain is \(0\leq x\leq10\) (option E), and the range should be a similar closed interval for \(y\). If we assume that the minimum distance is \(0\) and the maximum distance is \(10\) (for consistency), the range is \(0\leq y\leq10\), and the option that represents the range in the form \(s\leq y\leq t\) (option A) with \(s = 0\) and \(t = 10\) (but we have to go with the option selection).

Final Answers

• For the domain: \(\boldsymbol{E. \ 0\leq x\leq10}\)
• For the range: If we assume the minimum distance is \(0\) and maximum is \(10\) (matching the domain's upper limit for consistency), and the range is \(0\leq y\leq10\), and the option that represents the range in the form \(s\leq y\leq t\) (option A) with \(s = 0\) and \(t = 10\), but if we have to choose from the given options with the correct format, and assuming the problem's context, the range is represented by an inequality like \(0\leq y\leq10\) (and the corresponding option, if we take the first range option as \(0\leq y\leq10\), but we have to match the option. If we consider the options, and the motion of the car (starts, moves toward the wall (distance decreases), stops, moves back (distance increases), stops), the distance from the wall will have a minimum (when it is closest to the wall) and a maximum (when it is farthest from the wall). So the range is a closed interval for \(y\), so the answer for range (if we take the option with the correct format) is, for example, if the minimum distance is \(0\) and maximum is \(10\), the range is \(0\leq y\leq10\) (and the option would be the one with \(s\leq y\leq t\) with \(s = 0\) and \(t = 10\)).

(Note: The above solution is based on the assumption of a typical problem - setting where the domain and range have consistent upper limits for the purpose of illustration. The actual values may vary depending on the exact graph which is not fully visible, but the approach to identifying domain (input - time) and range (output - distance) as closed intervals is correct for this type of motion - related function.)