Answer :
To determine over which interval the function [tex]\( g(t) = -(t-1)^2 + 5 \)[/tex] has an average rate of change of zero, we need to calculate the average rate of change for each given interval and see where it equates to zero.
The formula for the average rate of change of a function [tex]\( g(t) \)[/tex] between two points [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] is:
[tex]\[ \frac{g(t_2) - g(t_1)}{t_2 - t_1} \][/tex]
Let's calculate the average rate of change for each interval step-by-step.
### Interval A: [tex]\( 1 \leq t \leq 4 \)[/tex]
1. Calculate [tex]\( g(1) \)[/tex]:
[tex]\[ g(1) = -(1-1)^2 + 5 = 5 \][/tex]
2. Calculate [tex]\( g(4) \)[/tex]:
[tex]\[ g(4) = -(4-1)^2 + 5 = -(3)^2 + 5 = -9 + 5 = -4 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(4) - g(1)}{4 - 1} = \frac{-4 - 5}{3} = \frac{-9}{3} = -3 \][/tex]
### Interval B: [tex]\( -4 \leq t \leq -3 \)[/tex]
1. Calculate [tex]\( g(-4) \)[/tex]:
[tex]\[ g(-4) = -((-4)-1)^2 + 5 = -(5)^2 + 5 = -25 + 5 = -20 \][/tex]
2. Calculate [tex]\( g(-3) \)[/tex]:
[tex]\[ g(-3) = -((-3)-1)^2 + 5 = -(4)^2 + 5 = -16 + 5 = -11 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(-3) - g(-4)}{-3 - (-4)} = \frac{-11 - (-20)}{-3 + 4} = \frac{-11 + 20}{1} = \frac{9}{1} = 9 \][/tex]
### Interval C: [tex]\( -2 \leq t \leq 0 \)[/tex]
1. Calculate [tex]\( g(-2) \)[/tex]:
[tex]\[ g(-2) = -((-2)-1)^2 + 5 = -(3)^2 + 5 = -9 + 5 = -4 \][/tex]
2. Calculate [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = -((0)-1)^2 + 5 = -(1)^2 + 5 = -1 + 5 = 4 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(0) - g(-2)}{0 - (-2)} = \frac{4 - (-4)}{0 + 2} = \frac{4 + 4}{2} = \frac{8}{2} = 4 \][/tex]
### Interval D: [tex]\( -2 \leq t \leq 4 \)[/tex]
1. Calculate [tex]\( g(-2) \)[/tex] (previously calculated as -4).
2. Calculate [tex]\( g(4) \)[/tex] (previously calculated as -4).
3. Average rate of change:
[tex]\[ \frac{g(4) - g(-2)}{4 - (-2)} = \frac{-4 - (-4)}{4 + 2} = \frac{-4 + 4}{6} = \frac{0}{6} = 0 \][/tex]
### Conclusion
The interval over which the function [tex]\( g(t) = -(t-1)^2 + 5 \)[/tex] has an average rate of change of zero is:
[tex]\[ \boxed{-2 \leq t \leq 4} \][/tex]
The formula for the average rate of change of a function [tex]\( g(t) \)[/tex] between two points [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] is:
[tex]\[ \frac{g(t_2) - g(t_1)}{t_2 - t_1} \][/tex]
Let's calculate the average rate of change for each interval step-by-step.
### Interval A: [tex]\( 1 \leq t \leq 4 \)[/tex]
1. Calculate [tex]\( g(1) \)[/tex]:
[tex]\[ g(1) = -(1-1)^2 + 5 = 5 \][/tex]
2. Calculate [tex]\( g(4) \)[/tex]:
[tex]\[ g(4) = -(4-1)^2 + 5 = -(3)^2 + 5 = -9 + 5 = -4 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(4) - g(1)}{4 - 1} = \frac{-4 - 5}{3} = \frac{-9}{3} = -3 \][/tex]
### Interval B: [tex]\( -4 \leq t \leq -3 \)[/tex]
1. Calculate [tex]\( g(-4) \)[/tex]:
[tex]\[ g(-4) = -((-4)-1)^2 + 5 = -(5)^2 + 5 = -25 + 5 = -20 \][/tex]
2. Calculate [tex]\( g(-3) \)[/tex]:
[tex]\[ g(-3) = -((-3)-1)^2 + 5 = -(4)^2 + 5 = -16 + 5 = -11 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(-3) - g(-4)}{-3 - (-4)} = \frac{-11 - (-20)}{-3 + 4} = \frac{-11 + 20}{1} = \frac{9}{1} = 9 \][/tex]
### Interval C: [tex]\( -2 \leq t \leq 0 \)[/tex]
1. Calculate [tex]\( g(-2) \)[/tex]:
[tex]\[ g(-2) = -((-2)-1)^2 + 5 = -(3)^2 + 5 = -9 + 5 = -4 \][/tex]
2. Calculate [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = -((0)-1)^2 + 5 = -(1)^2 + 5 = -1 + 5 = 4 \][/tex]
3. Average rate of change:
[tex]\[ \frac{g(0) - g(-2)}{0 - (-2)} = \frac{4 - (-4)}{0 + 2} = \frac{4 + 4}{2} = \frac{8}{2} = 4 \][/tex]
### Interval D: [tex]\( -2 \leq t \leq 4 \)[/tex]
1. Calculate [tex]\( g(-2) \)[/tex] (previously calculated as -4).
2. Calculate [tex]\( g(4) \)[/tex] (previously calculated as -4).
3. Average rate of change:
[tex]\[ \frac{g(4) - g(-2)}{4 - (-2)} = \frac{-4 - (-4)}{4 + 2} = \frac{-4 + 4}{6} = \frac{0}{6} = 0 \][/tex]
### Conclusion
The interval over which the function [tex]\( g(t) = -(t-1)^2 + 5 \)[/tex] has an average rate of change of zero is:
[tex]\[ \boxed{-2 \leq t \leq 4} \][/tex]