Respuesta :
[tex]\begin{gathered} \text{Given} \\ H(n)=\frac{3}{n+1}\text{ on }\lbrack2,4\rbrack \end{gathered}[/tex]
Recall the average rate of change
[tex]\text{Average Rate of Change}=\frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a}[/tex]Given the function H(n)
b = 4, a = 2.
Substitute the following given and we have the equation
[tex]\begin{gathered} \frac{\Delta y}{\Delta x}=\frac{H(4)-H(2)}{4-2} \\ \frac{\Delta y}{\Delta x}=\frac{(\frac{3}{4+1})-(\frac{3}{2+1})}{2} \\ \frac{\Delta y}{\Delta x}=\frac{(\frac{3}{5})-(\frac{3}{3})}{2} \\ \frac{\Delta y}{\Delta x}=\frac{\frac{3}{5}-1}{2} \\ \frac{\Delta y}{\Delta x}=\frac{\frac{3}{5}-\frac{5}{5}}{2} \\ \frac{\Delta y}{\Delta x}=\frac{\frac{-2}{5}}{2} \\ \frac{\Delta y}{\Delta x}=-\frac{2}{5}\cdot\frac{1}{2} \\ \frac{\Delta y}{\Delta x}=-\frac{\cancel{2}}{5}\cdot\frac{1}{\cancel{2}} \\ \frac{\Delta y}{\Delta x}=-\frac{1}{5} \end{gathered}[/tex]Therefore, the average rate of change of H(n) in the interval [2.4] is -1/5.
