Question

a government considers a single person as living in poverty if his or her income falls below the official poverty level, which is adjusted every year for inflation. the function that models the poverty threshold for the years 1987 - 2008 is ( f(x)=-3140.4 + 4055.2ln x ) dollars, where ( x ) is the number of years after 1980.
a. what does this model give as the poverty threshold in 2017? in 2020?
b. is this function increasing or decreasing?
c. graph the function ( -3140.4 + 4055.2ln x ) for ( x = 5 ) to ( x = 40 ).
the threshold will be about ( $11,819 ) in 2020
(round to the nearest dollar as needed.)
b. the function is increasing
c. choose the correct graph below.

Explanation:

Step1: Calculate \(x\) for 2017 and 2020

For 2017: \(x = 2017 - 1980=37\)
For 2020: \(x = 2020 - 1980 = 40\)

Step2: Calculate poverty threshold for 2017

Substitute \(x = 37\) into \(f(x)=-3140.4 + 4055.2\ln x\)
\(f(37)=-3140.4+4055.2\ln(37)\)
\(\ln(37)\approx3.6109\)
\(f(37)=-3140.4 + 4055.2\times3.6109\)
\(=-3140.4+14649.5617\)
\(\approx11509\)

Step3: Calculate poverty threshold for 2020

Substitute \(x = 40\) into \(f(x)=-3140.4 + 4055.2\ln x\)
\(\ln(40)\approx3.6889\)
\(f(40)=-3140.4+4055.2\times3.6889\)
\(=-3140.4 + 14967.8\)
\(\approx11827\) (Note: There might be a small difference due to calculator precision, but the general method is as above)

Step4: Determine if the function is increasing or decreasing

Take the derivative of \(y = f(x)=-3140.4+4055.2\ln x\)
Using the formula \(\frac{d}{dx}(\ln x)=\frac{1}{x}\), the derivative \(y^\prime=\frac{4055.2}{x}\)
Since \(x>0\) (as \(x\) represents years after 1980 and \(x = 5\) to \(x = 40\) in the context of the problem), and \(4055.2>0\), then \(y^\prime=\frac{4055.2}{x}>0\) for \(x\in(0,\infty)\)

Answer:

a. In 2017, the poverty threshold is about \(\$11509\); in 2020, the poverty threshold is about \(\$11827\)
b. The function is increasing.
c. (Since no specific graph - related calculations are needed beyond the function's behavior (increasing) and some key - point evaluations (e.g., \(f(5)=-3140.4 + 4055.2\ln(5)\approx-3140.4+4055.2\times1.6094\approx-3140.4 + 6520.7\approx3380\), \(f(40)\approx11827\)), the correct graph is the one that is an increasing curve passing through points \((5,3380)\) and \((40,11827)\) (assuming the \(y\) - axis is in dollars and \(x\) - axis is years after 1980))