Let's solve the problem step by step. ### Given Function \[ g(t) = - (t-1)^2 + 5 \] ### Step 1: **Recall the Average Rate of Change Formula** The average rate of change of \( g \) on the interval \([a, b]\) is: \[ \frac{g(b) - g(a)}{b - a} \] We want this to be **zero**, so: \[ g(b) - g(a) = 0 \implies g(a) = g(b) \] ### Step 2: **Check Each Interval** Let's check which interval gives \( g(a) = g(b) \). --- #### **Option (A):** \( -2 \leq t \leq 0 \) - \( a = -2 \), \( b = 0 \) - \( g(-2) = -((-2)-1)^2 + 5 = -(-3)^2 + 5 = -9 + 5 = -4 \) - \( g(0) = -((0)-1)^2 + 5 = -(-1)^2 + 5 = -1 + 5 = 4 \) - \( g(-2) \neq g(0) \) --- #### **Option (B):** \( -4 \leq t \leq -3 \) - \( a = -4 \), \( b = -3 \) - \( g(-4) = -((-4)-1)^2 + 5 = -(-5)^2 + 5 = -25 + 5 = -20 \) - \( g(-3) = -((-3)-1)^2 + 5 = -(-4)^2 + 5 = -16 + 5 = -11 \) - \( g(-4) \neq g(-3) \) --- #### **Option (C):** \( -2 \leq t \leq 4 \) - \( a = -2 \), \( b = 4 \) - \( g(-2) = -((-2)-1)^2 + 5 = -(-3)^2 + 5 = -9 + 5 = -4 \) - \( g(4) = -((4)-1)^2 + 5 = -(3)^2 + 5 = -9 + 5 = -4 \) - \( g(-2) = g(4) \) So the average rate of change is zero. --- #### **Option (D):** \( 1 \leq t \leq 4 \) - \( a = 1 \), \( b = 4 \) - \( g(1) = -((1)-1)^2 + 5 = -(0)^2 + 5 = 0 + 5 = 5 \) - \( g(4) = as above = -4 \) - \( g(1) \neq g(4) \) --- ### **Final Answer** \[ \boxed{(-2 \leq t \leq 4)} \] **So, Option C is correct.** **Explanation:** The average rate of change is zero when the function has the same value at both endpoints of the interval. For \( t = -2 \) and \( t = 4 \), \( g(-2) = g(4) = -4 \). Thus, the average rate of change on this interval is zero.

The average rate of change of a function over an interval is zero when the function's values at the endpoints are equal. This indicates the function attains the same value at both points, implying the net change over that interval is zero. Therefore, the interval where the function's average rate of change is zero corresponds to where \(g(a) = g(b)\).