## Calculus 10th Edition

For all three cases in this exercise (average rate of change and instantaneous rates of change for both endpoints $[1,2]$) the answer is 4.
Average rate of change is given by the formula $$\frac{f(t_2) - f(t_1)}{t_2 - t_1}$$ for points $(t_1, f(t_1))$ and $(t_2, f(t_2))$ respectively. In this case, for endpoints $[1,2]$: $$f(t) = 4t +5$$ $$f(1) = 4(1) + 5 = 9$$ $$f(2) = 4(2) + 5 = 13$$ and the average rate of change is $$Average Rate = \frac{13 - 9}{2-1} = \frac{4}{1} = 4$$ For the instantaneous rates of change, we need the first derivative: $$f(t) = 4t + 5$$ $$f'(t) = 4t^{1-0} + 0$$ $$f'(t) = 4$$ which shows that the answer is $4$ for all possible values of $t$.