Answer :
To determine over which interval the function \( h(t) = (t + 3)^2 + 5 \) has a negative average rate of change, we need to find the average rate of change over each of the given intervals and identify the one where this rate is negative.
The average rate of change of the function \( h(t) \) over an interval \([a, b]\) is given by:
[tex]\[ \frac{h(b) - h(a)}{b - a} \][/tex]
Let's examine each interval one by one:
### Interval 1: \(-2 \leq t \leq 0\)
- \( a = -2 \)
- \( b = 0 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-2) = ((-2 + 3)^2) + 5 = 1^2 + 5 = 6 \][/tex]
[tex]\[ h(0) = ((0 + 3)^2) + 5 = 3^2 + 5 = 14 \][/tex]
Average rate of change:
[tex]\[ \frac{h(0) - h(-2)}{0 - (-2)} = \frac{14 - 6}{2} = \frac{8}{2} = 4 \][/tex]
### Interval 2: \(1 \leq t \leq 4\)
- \( a = 1 \)
- \( b = 4 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(1) = ((1 + 3)^2) + 5 = 4^2 + 5 = 16 + 5 = 21 \][/tex]
[tex]\[ h(4) = ((4 + 3)^2) + 5 = 7^2 + 5 = 49 + 5 = 54 \][/tex]
Average rate of change:
[tex]\[ \frac{h(4) - h(1)}{4 - 1} = \frac{54 - 21}{3} = \frac{33}{3} = 11 \][/tex]
### Interval 3: \(-4 \leq t \leq -3\)
- \( a = -4 \)
- \( b = -3 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-4) = ((-4 + 3)^2) + 5 = (-1)^2 + 5 = 1 + 5 = 6 \][/tex]
[tex]\[ h(-3) = ((-3 + 3)^2) + 5 = 0^2 + 5 = 5 \][/tex]
Average rate of change:
[tex]\[ \frac{h(-3) - h(-4)}{-3 - (-4)} = \frac{5 - 6}{1} = \frac{-1}{1} = -1 \][/tex]
### Interval 4: \(-3 \leq t \leq 4\)
- \( a = -3 \)
- \( b = 4 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-3) = ((-3 + 3)^2) + 5 = 0^2 + 5 = 5 \][/tex]
[tex]\[ h(4) = ((4 + 3)^2) + 5 = 7^2 + 5 = 49 + 5 = 54 \][/tex]
Average rate of change:
[tex]\[ \frac{h(4) - h(-3)}{4 - (-3)} = \frac{54 - 5}{7} = \frac{49}{7} = 7 \][/tex]
From the calculations, we see that the interval where the average rate of change is negative is:
\[
\text{Interval 3: } \boxed{-4 \leq t \leq -3}
The average rate of change of the function \( h(t) \) over an interval \([a, b]\) is given by:
[tex]\[ \frac{h(b) - h(a)}{b - a} \][/tex]
Let's examine each interval one by one:
### Interval 1: \(-2 \leq t \leq 0\)
- \( a = -2 \)
- \( b = 0 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-2) = ((-2 + 3)^2) + 5 = 1^2 + 5 = 6 \][/tex]
[tex]\[ h(0) = ((0 + 3)^2) + 5 = 3^2 + 5 = 14 \][/tex]
Average rate of change:
[tex]\[ \frac{h(0) - h(-2)}{0 - (-2)} = \frac{14 - 6}{2} = \frac{8}{2} = 4 \][/tex]
### Interval 2: \(1 \leq t \leq 4\)
- \( a = 1 \)
- \( b = 4 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(1) = ((1 + 3)^2) + 5 = 4^2 + 5 = 16 + 5 = 21 \][/tex]
[tex]\[ h(4) = ((4 + 3)^2) + 5 = 7^2 + 5 = 49 + 5 = 54 \][/tex]
Average rate of change:
[tex]\[ \frac{h(4) - h(1)}{4 - 1} = \frac{54 - 21}{3} = \frac{33}{3} = 11 \][/tex]
### Interval 3: \(-4 \leq t \leq -3\)
- \( a = -4 \)
- \( b = -3 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-4) = ((-4 + 3)^2) + 5 = (-1)^2 + 5 = 1 + 5 = 6 \][/tex]
[tex]\[ h(-3) = ((-3 + 3)^2) + 5 = 0^2 + 5 = 5 \][/tex]
Average rate of change:
[tex]\[ \frac{h(-3) - h(-4)}{-3 - (-4)} = \frac{5 - 6}{1} = \frac{-1}{1} = -1 \][/tex]
### Interval 4: \(-3 \leq t \leq 4\)
- \( a = -3 \)
- \( b = 4 \)
Calculate \( h(a) \) and \( h(b) \):
[tex]\[ h(-3) = ((-3 + 3)^2) + 5 = 0^2 + 5 = 5 \][/tex]
[tex]\[ h(4) = ((4 + 3)^2) + 5 = 7^2 + 5 = 49 + 5 = 54 \][/tex]
Average rate of change:
[tex]\[ \frac{h(4) - h(-3)}{4 - (-3)} = \frac{54 - 5}{7} = \frac{49}{7} = 7 \][/tex]
From the calculations, we see that the interval where the average rate of change is negative is:
\[
\text{Interval 3: } \boxed{-4 \leq t \leq -3}
