G(t)=−(t−1)^2+5

over which interval does g have an average rate of change of zero?

choose 1

(choice a)

1≤t≤4

(choice b)

−4≤t≤−3

(choice c)

−2≤t≤0

(choice d)

−2≤t≤4

In the interval of - 2 ≤ t ≤ 4, the average rate of change is zero.

Step-by-step explanation:

The function is given to be  ............ (1)

(i) Now, for 1 ≤ t ≤ 4, we will put t = 1 and t = 4.

So, g(1) = 5, g(4) = - 4.  {From equation (1)}

Hence, g(1) ≠ g(4)

(ii) Now, for - 4 ≤ t ≤ - 3, we will put t = - 4 and t = - 3.

So, g(- 4) = - 20, g(- 3) = - 11.  {From equation (1)}

Hence, g(- 4) ≠ g(- 3)

(iii) Now, for -2 ≤ t ≤ 0, we will put t = - 2 and t = 0.

So, g(- 2) = - 4, g(0) = 4.  {From equation (1)}

Hence, g(- 2) ≠ g(0)

(iv) Now, for - 2 ≤ t ≤ 4, we will put t = - 2 and t = 4.

So, g(- 2) = - 4, g(4) = - 4.  {From equation (1)}

Hence, g(- 2) = g(4)

Therefore, in the interval of - 2 ≤ t ≤ 4, the average rate of change is zero. (Answer)

it could be 1,3 and more others to create a cylinder.

step-by-step explanation: