Answer
(a) $f(0) = 88$
(b) $f(2) = 72, f(4) = 64, f(6) = 59, f(8) = 55, f(10) = 52, f(12) = 50$
(c) See graph; by 350mo after the original exam, students would seem to remember nothing.
Work Step by Step
$$f(t) = 88 - 15\ln(t+1); 0 \leq t \leq 12$$
(a) Since $f(t)$ models the $average$ score, what we need for the $original$ exam is the value $t = 0$, since it was at the very beginning: $$f(0) = 88 - 15\ln(0+1)$$ $$f(0) = 88 - 15\ln(1) = 88-15(0) = 88$$
(b) $$f(2) = 88 - 15\ln(2+1) = 88-15\ln(3) \approx 72$$ $$f(4) = 88 - 15\ln(4+1) = 88 - 15\ln(5) \approx 64$$ $$f(6) = 88 - 15\ln(6+1) = 88 - 15\ln(7) \approx 59$$ $$f(8) = 88 - 15\ln(8+1) = 88 - 15\ln(9) \approx 55$$ $$f(10) = 88 - 15\ln(10+1) = 88 - 15\ln(11) \approx 52$$ $$f(12) = 88 - 15\ln(12+1) = 88 - 15\ln(13) \approx 50$$
(c) By approximately 350 months, students seem to retain no knowledge of the original material.
