Out-of-pocket spending in a country for health care increased between 2004 and 2008. The function f(x) = 2572 0.0359% models average annual expenditures perhousehold, in dollars. In this model, x represents the year where x = 0 corresponds to 2004(a) Estimate out-of-pocket household spending on health care in 2008(1) Determine the year when spending reached $2807 per household(a) The total expenditures per household in the year 2008 were approximately $(Round to the nearest dollar as needed)(b) During the year spending reached $2807 per household.(Round down to the nearest year)7

And x = 0 corresponds to 2004, we can estimate that the year 2008 corresponds to x = 4, then, if we put x = 4 in our function we will have the estimated out-of-pocket household spending on health care in 2008. Therefore

[tex]\begin{gathered} f(4)=2572e^{0.0359\cdot4} \\ \\ f(4)=2572e^{0.1436} \\ \\ f(4)=2969.17401995 \end{gathered}[/tex]

If we round it to the nearest dollar we will have

[tex]f(4)=2969[/tex]

Final answer:

a) The total expenditures per household in the year 2008 were approximately $2969

For the second item:

Now we have the value of f(x) and we want to find which value of x satisfies the equation:

[tex]2807=2572e^{0.0359x}[/tex]

To solve that equation we will need to apply the natural logarithm on both sides and remember that:

[tex]\begin{gathered} \ln (e)=1 \\ \\ \ln (e^x)=x \end{gathered}[/tex]

Then, doing the ln both sides we have

[tex]\begin{gathered} \ln (2807)=\ln (2572e^{0.0359x}) \\ \\ \ln (2807)=\ln (2572)+\ln (e^{0.0359x}) \\ \\ \ln (2807)=\ln (2572)+0.0359x\ln (e) \\ \\ \ln (2807)=\ln (2572)+0.0359x \end{gathered}[/tex]

Now we have a "linear equation" and we can solve it for x, it will be

[tex]\begin{gathered} \ln (2807)=\ln (2572)+0.0359x \\ \\ 0.0359x=\ln (2807)-\ln (2572) \\ \\ x=\frac{\ln(2807)-\ln(2572)}{0.0359}=2.435\text{ year} \end{gathered}[/tex]

Rounding it to the nearest year, it will be 2 years.

Final answer:

During the year 2006 spending reached $2807 per household.