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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 = ________

User Forsberg
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1 Answer

4 votes
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

User Ramanr
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