It is given the function $f:\mathbb{R}\rightarrow \mathbb{R}$ fow which $f(1)=1$ and for all $x\in\mathbb{R}$ satisfied $f(x+5)\geq f(x)+5$ and $f(x+1)\leq f(x)+1$ If $g(x)=f(x)-x+1$ then find $g(2016)$ .
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Tags: functional equation
adityaguharoy
09.01.2017 18:56
dangerousliri wrote: It is given the function $f:\mathbb{R}\rightarrow \mathbb{R}$ fow which $f(1)=1$ and for all $x\in\mathbb{R}$ satisfied property 1 $f(x+5)\geq f(x)+5$ and property 2 $f(x+1)\leq f(x)+1$ If $g(x)=f(x)-x+1$ then find $g(2016)$ . Note that : From property 2 we get : $f(x+2) \le f(x+1)+1 \le f(x)+1+1=f(x)+2$ $f(x+3) \le f(x+2)+1 \le f(x)+2+1=f(x)+3$ $f(x+4) \le f(x+3)+1 \le f(x)+3+1=f(x)+4$ $f(x+5) \le f(x+4)+1 \le f(x)+4+1=f(x)+5$ But, given $f(x+5) \ge f(x)+5$ So, we must have $f(x+5)=f(x)+5$ Thus, $f(6)=6,f(11)=11,f(16)=16,f(21)=21,............,f(5k+1)=5k+1$ (this is easy to show by induction) and thus $f(2016)=2016$ and thus, $g(2016)=1$
jicko
22.12.2019 08:30
$
f(x+5)>=f(x)=5
and at the same time : f(x+1)<=f(x)+1 :f(x+2)<=f(x+1)+1<=f(x)+2
thus is general:f(x+n)<=f(x)+n
thus f(x)+5<=f(x+5)<=f(x)+5:thus f(x)=x for all x, x>5:thus g(x)=1 for all x,x>5$
tommy2007
24.02.2020 12:11
jicko wrote:
$
f(x+5)>=f(x)=5
and at the same time : f(x+1)<=f(x)+1 :f(x+2)<=f(x+1)+1<=f(x)+2
thus is general:f(x+n)<=f(x)+n
thus f(x)+5<=f(x+5)<=f(x)+5:thus f(x)=x for all x, x>5:thus g(x)=1 for all x,x>5$
man, it is so squeezed.