Question

Find the value of constant k (or indicate that no such value of k exists), so that the average rate of change in F(t) = kt^2 on the interval [-1, 2] is equal to 1. K = ________

Answer 1

Answer:

k = 1

Step-by-step explanation:

Given the function:

`f(t)=kt^2`

The average rate of change is given by the formula:

`R=(f(b)-f(a))/(b-a)`

where a = -1, b = 2

`\begin{gathered} R=(2^2k-(-1)^2k)/(2-(-1)) \\ \\ =(4k-k)/(2+1) \\ \\ =(3k)/(3)=k \end{gathered}`

Now, the rate of change has been given to be 1, so

`R=k=1`