<h3>Given</h3>
The values of two houses (in thousands of dollars)
![\left[\begin{array}{c|cccc}\text{year}&0&1&2&3\\\text{value 1}&286&294.58&303.4174&312.51992\\\text{value 2}&286&295&304&313\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Ccccc%7D%5Ctext%7Byear%7D%260%261%262%263%5C%5C%5Ctext%7Bvalue%201%7D%26286%26294.58%26303.4174%26312.51992%5C%5C%5Ctext%7Bvalue%202%7D%26286%26295%26304%26313%5Cend%7Barray%7D%5Cright%5D)
<h3>Find</h3>
A) the nature of the function, linear or exponential, that can be used to model the value after x years
B) the actual function f(x) that can be used in each case
C) f(25) for each house. Is there a significant difference?
<h3>Solution</h3>
A) The oddball numbers give you a clue immediately that the value of house 1 will be best modeled by an exponential function.
The value of house 2 is increasing steadily at 9,000 per year, so is modeled by a linear function.
B) The ratio of values from a given year to the year before for house 1 is
... 294.58/286 = 1.03
A check for other years reveals the same ratio, so the exponential function can be written for house 1 as
... f(x) = 286·1.03^x . . . . . value of house 1
In part A we determined the year-to-year difference in value for house 2 is 9,000. That is the slope of the linear function. Then (in thousands), that function is
... f(x) = 286 +9x . . . . . value of house 2
C) After 25 years, the house values are (in thousands of dollars)
The value of house 1 has more than doubled in the same time that the value of house 2 has increased by about 79%. This is a significant difference.
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An exponential function will always outperform a linear function over a long enough time period.
