It is given that f(x) is a function defined on R, satisfying f(1) =
1, and for any x ∈ R,
f(x+5)>f(x)+4 and f(x+1)<f(x)+2
.
If g(x)=f(x)+1x evaluate g(2018).
For the simplest recursion, where f(x+1)=f(x)+k, k can be anywhere from just above 4/5 to just below 2.
If k = .8, g(2018) would equal  402.4
If k = 2, g(2018) would equal 2018
If k=1, g(2018) would be 1.

Posted by Charlie
on 20180408 11:48:35 