Question

the function ( f(x)=60 e^{-0.4 x}+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, ( 100.0 % ) of the information is remembered.
(round to one decimal place as needed.)
b. after one week, ( 80.2 % ) of the information is remembered.
(round to one decimal place as needed.)
c. after four weeks, ( 52.1 % ) 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: Substitute \(x = 52\) into the function

We have the function \(f(x)=60e^{-0.4x}+40\). Substituting \(x = 52\) gives \(f(52)=60e^{-0.4\times52}+40\).
First, calculate the exponent: \(-0.4\times52=-20.8\). So the function becomes \(f(52)=60e^{- 20.8}+40\).

Step2: Evaluate \(e^{-20.8}\)

We know that \(e^{-a}=\frac{1}{e^{a}}\). Since \(e^{20.8}\) is a very large number. \(e^{-20.8}\approx0\) (because as \(a\) gets larger, \(e^{-a}\) approaches \(0\)).
Then \(60e^{-20.8}\approx60\times0 = 0\).

Step3: Calculate \(f(52)\)

\(f(52)=60e^{-20.8}+40\approx0 + 40\)

Answer:

\(40.0\)