f(x^2*2016x) = f(x)x+2016 Then f(2017) = ?
Mar 29, 2020 #2 HallsofIvy Science Advisor Homework Helper 43,021 973 Isn't "x^2*2016x" the same as "2016x^3"?
Mar 29, 2020 #3 skeeter 1,104 1 similar problem from another site had $f(x^2-2016x) = f(x) \cdot x + 2016$ https://mathhelpforum.com/threads/functions-problem.285171/#post-954658 very good hint was given to determine $f(2017)$
similar problem from another site had $f(x^2-2016x) = f(x) \cdot x + 2016$ https://mathhelpforum.com/threads/functions-problem.285171/#post-954658 very good hint was given to determine $f(2017)$
Apr 21, 2020 #4 HallsofIvy Science Advisor Homework Helper 43,021 973 Thanks, skeeter. That sounds more reasonable. f(x^2- 2016x)= f(2017) if x^2- 2016x= 2017. That is the same as x^2- 2016x- 2017= (x+ 1)(x- 2017)= 0 so either x= -1 or x= 2017. So either f(2017)= -1(f(-1))+ 2016 or f(2017)= 2017 f(2017)+ 2016. For the latter, -2016= 2016 f(2017) so f(2017)= -1.
Thanks, skeeter. That sounds more reasonable. f(x^2- 2016x)= f(2017) if x^2- 2016x= 2017. That is the same as x^2- 2016x- 2017= (x+ 1)(x- 2017)= 0 so either x= -1 or x= 2017. So either f(2017)= -1(f(-1))+ 2016 or f(2017)= 2017 f(2017)+ 2016. For the latter, -2016= 2016 f(2017) so f(2017)= -1.