Question

the function $f(x) = 60e^{-0.3x} + 40$ describes the percentage of information, $f(x)$, that a particular person remembers $x$ weeks after learning the information.
a. substitute 0 for $x$ and, without using a calculator, find the percentage of information remembered at the moment it is first learned.
b. substitute 1 for $x$ and find the percentage of information that is remembered after 1 week.
c. find the percentage of information that is remembered after 4 weeks.
d. find the percentage of information that is remembered after one year (52 weeks).
a. at the moment it is first learned, \\(\square\\)% of the information is remembered. (round to one decimal place as needed.)
b. after one week, \\(\square\\)% of the information is remembered. (round to one decimal place as needed.)
c. after four weeks, \\(\square\\)% of the information is remembered. (round to one decimal place as needed.)
d. after one year, \\(\square\\)% of the information is remembered. (round to one decimal place as needed.)

Explanation:

Step1: Solve part a (x=0)

The function is \( f(x) = 60e^{-0.3x} + 40 \). Substitute \( x = 0 \):
\( f(0) = 60e^{-0.3(0)} + 40 = 60e^{0} + 40 = 60(1) + 40 = 100 \).

Step2: Solve part b (x=1)

Substitute \( x = 1 \) into \( f(x) \):
\( f(1) = 60e^{-0.3(1)} + 40 \approx 60(0.7408) + 40 \approx 44.448 + 40 = 84.448 \), round to one decimal: \( 84.4 \).

Step3: Solve part c (x=4)

Substitute \( x = 4 \):
\( f(4) = 60e^{-0.3(4)} + 40 = 60e^{-1.2} + 40 \approx 60(0.3012) + 40 \approx 18.072 + 40 = 58.072 \), round to one decimal: \( 58.1 \).

Step4: Solve part d (x=52)

Substitute \( x = 52 \):
\( f(52) = 60e^{-0.3(52)} + 40 = 60e^{-15.6} + 40 \approx 60(0) + 40 = 40.0 \) (since \( e^{-15.6} \) is extremely small, approximately 0).

Answer:

a. \( 100.0 \)
b. \( 84.4 \)
c. \( 58.1 \)
d. \( 40.0 \)