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

Explanation:

Given the function:

$f(t)=kt^2$

The average rate of change is given by the formula:

$R=\frac{f(b)-f(a)}{b-a}$

where a = -1, b = 2

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

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

$R=k=1$