To determine the range of the function [tex]\( c(x) = -10x + 150 \)[/tex], we first need to understand what the function represents and identify the possible values it can take.
The linear function [tex]\( c(x) = -10x + 150 \)[/tex] models the number of pieces of candy remaining, [tex]\( c(x) \)[/tex], after [tex]\( x \)[/tex] days. Here:
- [tex]\( x \)[/tex] represents the number of days,
- [tex]\( c(x) \)[/tex] represents the remaining pieces of candy.
### Key Points:
1. Initial Value: At [tex]\( x = 0 \)[/tex] days, [tex]\( c(0) = -10(0) + 150 = 150 \)[/tex]. This means at the start, each member has 150 pieces of candy.
2. Rate of Decrease: The term [tex]\( -10x \)[/tex] shows that the number of pieces of candy decreases by 10 pieces each day.
3. Minimum Value: When all the candy is sold out, the number of pieces of candy remaining [tex]\( c(x) \)[/tex] will be 0. We need to find the value of [tex]\( x \)[/tex] when [tex]\( c(x) = 0 \)[/tex]:
Set the equation to 0:
[tex]\[
0 = -10x + 150
\][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[
10x = 150
\][/tex]
[tex]\[
x = 15
\][/tex]
This calculation shows that it takes 15 days to sell all 150 pieces of candy.
### Range:
- Maximum Value: The maximum number of pieces of candy remaining is 150 (at day 0).
- Minimum Value: The minimum number of pieces of candy remaining is 0 (at day 15).
Therefore, the range of the function [tex]\( c(x) = -10x + 150 \)[/tex] is from 0 pieces to 150 pieces. This means [tex]\( c(x) \)[/tex] can take any value between 0 and 150, inclusive.
### Conclusion:
The correct range of the function is:
[tex]\[
[0,150]
\][/tex]
Hence, the correct answer is:
[tex]\[
[0,150]
\][/tex]
