Question

Find the average rate of change of the following function from t = 1 to t=2.5h(t) = 148 – 16t

Answer 1

The average rate of change of the function from t=1 to t=2.5 is given by:

`(h(2.5)-h(1))/(2.5-1)=(h(2.5)-h(1))/(1.5)`

It is given that:

`\begin{gathered} h(t)=148-16t \\ h(2.5)=148-16*2.5=108 \\ h(1)=148-6=142 \end{gathered}`

Substitute the values to get:

`(h(2.5)-h(1))/(1.5)=(108-142)/(1.5)=(-68)/(3)\approx-22.6667`

Hence the rate of change is -22.6667.