Question

h(t)=(t+3)^2+5 What is the average rate of change of h over the interval -5

Answer 1

$h(t) = {(t + 3)}^{2} + 5 \\ h(t) = {t}^{2} + 6t + 9 + 5 \\ h(t) = {t}^{2} + 6t + 14$

average rate of change

$= \frac{dh(t)}{dt} \\ = \frac{d( {t}^{2} + 6t + 14) }{dt} \\ = \frac{d ({t}^{2}) }{dt} + \frac{d(6t)}{dt} + \frac{d(14)}{dt} \\ = 2{t}^{2 - 1} + 6 \times 1 {t}^{1 - 1} + 0 \\ = 2t + 6 {t}^{0} \\ = 2t + 6$

at the interval of 5 seconds, t = 5

hence, average rate of change at t = 5

= 2(5) + 6

= 16