Question

Find the value of constant k so that the average rate of change in f(t) on the indicated interval is the value provided, or indicate that no such value of k exists F(t) = kt^2; ARC = 1.5 on [-1, 2]

Answer 1

The averege rate of f(x) on the interval [a, b] is

$\text{Ave}\mathrm{}=\frac{f(b)-f(a)}{b-a}$

Since the given interval is [-1, 2], then

a = -1, b = 2

Since the function is

$f(t)=kt^2$

Since the average is 1.5

Substitute them in the rule above

$\begin{gathered} 1.5=\frac{k(2)^2-k(-1)^2}{2--1} \\ 1.5=\frac{4k-k}{2+1} \\ 1.5=\frac{3k}{3} \\ 1.5=k \end{gathered}$

The value of k is 1.5