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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?.

User Pkqk
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slope = m = \cfrac{rise}{run} \implies \cfrac{ f(x_2) - f(x_1)}{ x_2 - x_1}\impliedby \begin{array}{llll} average~rate\\ of~change \end{array}\\\\[-0.35em] \rule{34em}{0.25pt}\\\\ f(x)=18000(0.52)^x \qquad \begin{cases} x_1=4\\ x_2=8 \end{cases}\implies \cfrac{f(8)-f(4)}{8-4} \\\\\\ \cfrac{[18000(0.52)^8]~~ - ~~[18000(0.52)^4]}{4}~~\approx ~~\cfrac{-1219.86}{4}~~ \approx ~~ -304.97

notice, the rate is negative, because the car is losing value as the time moves on.

User Annie Sheikh
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