MATH SOLVE

5 months ago

Q:
# The function f(x)=18000(0.52)x represents the value in dollars of a vehicle x years after it has been purchased new.What is the average rate of change in value per year between years 4 and 8? −$18000/year −$1219.87/year−$304.97/year−$0.52/year

Accepted Solution

A:

Answer: −$304.97/year

Step-by-step explanation:We know that the rate of change of function=[tex]\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]Thus, for the required situation the rate of change of function per year between years 4 and 8=[tex]\frac{f(8)-f(4)}{8-4}\\\\=\frac{18000(0.52)^8-18000(0.52)^4}{4}\\\\=\frac{18000[(0.52)^8-(0.52)^4]}{4}\\\\=\frac{18000[0.00534-0.07311]}{4}\\\\=\frac{18000\times-0.06777}{4}\\\\=\frac{-1219.86}{4}=-304.97[/tex]Hence the average rate of change in value per year between years 4 and 8= -$304.97 per year.

Step-by-step explanation:We know that the rate of change of function=[tex]\frac{f(x_2)-f(x_1)}{x_2-x_1}[/tex]Thus, for the required situation the rate of change of function per year between years 4 and 8=[tex]\frac{f(8)-f(4)}{8-4}\\\\=\frac{18000(0.52)^8-18000(0.52)^4}{4}\\\\=\frac{18000[(0.52)^8-(0.52)^4]}{4}\\\\=\frac{18000[0.00534-0.07311]}{4}\\\\=\frac{18000\times-0.06777}{4}\\\\=\frac{-1219.86}{4}=-304.97[/tex]Hence the average rate of change in value per year between years 4 and 8= -$304.97 per year.