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Advanced Math / Nonlinear functions Difficulty: Hard

$f (x) = 5,470 {(0.64)}^{\frac{x}{12}}$

The function $f$ gives the value, in dollars, of a certain piece of equipment after $x$ months of use. If the value of the equipment decreases each year by $p %$ of its value the preceding year, what is the value of $p$ ?

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Explanation

Choice C is correct. For a function of the form $f (x) = a {(r)}^{\frac{x}{k}}$ , where $a$ , $r$ , and $k$ are constants and $r less than 1$ , the value of $f (x)$ decreases by $100 left parenthesis 1 minus r right parenthesis percent sign$ for every increase of $x$ by $k$ . In the given function, $a equals 5,470$ , $r equals 0.64$ , and $k equals 12$ . Therefore, for the given function, the value of $f (x)$ decreases by $100 left parenthesis 1 minus 0.64 right parenthesis percent sign$ , or $36 percent sign$ , for every increase of $x$ by $12$ . Since $f (x)$ represents the value, in dollars, of the equipment after $x$ months of use, it follows that the value of the equipment decreases every $12$ months by $36 percent sign$ of its value the preceding $12$ months. Since there are $12$ months in a year, the value of the equipment decreases each year by $36 percent sign$ of its value the preceding year. Thus, the value of $p$ is $36$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect and may result from conceptual or calculation errors.

Choice D is incorrect and may result from conceptual or calculation errors.