Solution:
Given that Juan bought a car for $32,500, and the function f(x) expressed as
[tex]\begin{gathered} f(x)=32500(0.835)^x \\ \text{where} \\ x\Rightarrow\text{ number of years} \end{gathered}[/tex]
is the value of the car in any given year as shown in the table below:
From the above table, Juan bought the car in 2009.
Thus, to evaluate the approximate value of the car in 2020,
step 1: Evalaute the value of x.
In 2020, the value of x from 2009 will be
[tex]number\text{ of years betw}een\text{ 2009 and 2020 =11}[/tex]
thus,
[tex]x=11[/tex]
step 2: substitute 11 for the value of x into the f(x) function.
thus,
[tex]\begin{gathered} f(x)=32500(0.835)^x \\ \text{where x=11} \\ \Rightarrow f(x)=32500(0.835)^{11} \\ =4471.308047 \end{gathered}[/tex]
Hence, the approximate value of the car in 2020 is
[tex]\$4480[/tex]
The correct option is B.
