[tex]\begin{gathered} f(x)=55\cdot(0.7)^x \\ \end{gathered}[/tex]
Explanation
exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate., it is defined by the expression
[tex]\begin{gathered} f(x)=ab^x \\ where \\ a\text{ is the initial amount} \\ b\text{ is the decay factor ( b}<1) \\ x\text{ is the time} \end{gathered}[/tex]
so
Step 1
Let
[tex]\begin{gathered} \text{ Initial value=a=}55 \\ decay\text{ factor}=b=0.7 \\ \end{gathered}[/tex]
now,replace
[tex]\begin{gathered} f(x)=ab^x \\ f(x)=55\cdot(0.7)^x \end{gathered}[/tex]
so, the model/ function is
[tex]f(x)=55\cdot(0.7)^x[/tex]
Step 2
rate of change, it is geiven by:
[tex]\text{rate}=\frac{f(b)-f(a)}{b-a}[/tex]
let
[tex]\begin{gathered} a=1 \\ b=3 \end{gathered}[/tex]
replace
[tex]\begin{gathered} f(x)=55\cdot(0.7)^x \\ f(1)=55\cdot(0.7)^1 \\ f(1)=38.5 \\ f(3)=55\cdot(0.7)^3 \\ f(3)=18.865 \end{gathered}[/tex]
therefore
[tex]\begin{gathered} \text{rate}=\frac{f(b)-f(a)}{b-a} \\ rate=\frac{18.865-38.5}{3-1}=-9.8175 \end{gathered}[/tex]
we can conclude
[tex]\begin{gathered} \text{the average rate of change over 1}\leq x\leq3 \\ is \\ -9.817 \end{gathered}[/tex]
Step 3
over
[tex]4\leq x\leq8[/tex]
Let
[tex]\begin{gathered} a=4 \\ b=8 \\ f(4)=55\cdot(0.7)^4 \\ f(4)=13.2055 \\ f(8)=55\cdot(0.7)^8 \\ f(8)=3.17064055 \end{gathered}[/tex]
replace
[tex]\begin{gathered} \text{rate}=\frac{f(b)-f(a)}{b-a} \\ rate=\frac{3.17-13.20}{8-4}=-2.509 \\ rate=-2.509 \end{gathered}[/tex]
I hope this helps you
