Answer:
[tex]a^{\prime}_{\text{avg}}(t)=\frac{-1}{13(h+13)}[/tex]Explanation:
The average rate of change of a function a(t) is defined as
[tex]a^{\prime}_{\text{avg}}(t)=\frac{a(t+h)-a(t)}{h}[/tex]Now in our case, we have
[tex]a(t)=\frac{1}{t+4}[/tex]therefore,
[tex]a(9+h)=\frac{1}{9+h+4}=\frac{1}{h+13}[/tex]and
[tex]a(9)=\frac{1}{9+4}=\frac{1}{13}[/tex]Hence,
[tex]a^{\prime}_{\text{avg}}(t)=\frac{\frac{1}{h+13}-\frac{1}{13}}{h}[/tex]which we rewrite to get
[tex]a^{\prime}_{\text{avg}}(t)=\frac{1}{h}(\frac{1}{h+13}-\frac{1}{13})[/tex][tex]\Rightarrow\frac{1}{h}(\frac{13}{13(13+h)}-\frac{h+13}{13(h+13)})[/tex][tex]=\frac{1}{h}(\frac{13-(h+13)}{13(h+13)})[/tex][tex]=\frac{1}{h}(\frac{-h}{13(h+13)})[/tex][tex]=\frac{-1}{13(h+13)}[/tex]Hence,
[tex]\boxed{a^{\prime}_{\text{avg}}(t)=\frac{-1}{13(h+13)}}[/tex]which is our answer!