Question

Complete the average rate of change (ARC) for the function H(n) = 5/ n + 1 On the interval [2, 10] ARC[2, 10] = ________

Answer 1

$H(n)=\frac{5}{n+1}$

Averate rate of change:

$\begin{gathered} \lbrack a,b\rbrack \\ \\ ARC=\frac{H(b)-H(a)}{b-a} \end{gathered}$

For the given interval:

$\begin{gathered} \text{ARC}_{\lbrack2,10\rbrack}=\frac{H(10)-H(2)}{10-2} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=\frac{\frac{5}{10+1}-\frac{5}{2+1}}{8} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=\frac{\frac{5}{11}-\frac{5}{3}}{8} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=\frac{\frac{15-55}{33}}{8} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=\frac{-\frac{40}{33}}{8} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=\frac{-40}{33\cdot8} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=-\frac{40}{264} \\ \\ \text{ARC}_{\lbrack2,10\rbrack}=-\frac{5}{33}=-0.\bar{15} \end{gathered}$

Then, the averate rate of change of the given function in the interval {2,10} is -5/33 (or -0.15)