$f(1)=1$, $f(x+5) \ge f(x) + 5$, $f(x+1) \le f(x) + 1$, and $g(x) = f(x) + 1 - x$. Which of the folowing statement are true? If $f(1) = 1$, and 
$$f(x+5) \ge f(x) + 5$$, 
$$f(x+1) \le f(x) + 1$$, 
$$g(x) = f(x) + 1 - x$$. Which statements are true?


*

*$f(x+y) = f(x) + y$, for all $x,$ real

*$f(2016)=2017$

*$g(x) \le f(x)$, for all $x$ real

*$g(2016) = 2$

Attempt:
Notice that the 3rd statement is incorrect, because if $1-x > 0$ then $g(x) > f(x)$.
Now $f(0+1) = 1 \le f(0) + 1$, so $f(0)$ must be $0$. Notice also that $f(2) \le 2$ and $f(x) \le x$ for $x$ integers. Also 
$$f(0+5) \ge f(0) + 5 = 5$$
and
$$f(4+1)=f(5) \le f(4) + 1 \le 5$$
so we must have $f(5)=5$. Notice that this is true for all integers, $f(x)=x$.
So the 2nd statement is also incorrect.
Also $g(2016) = f(2016) - 2015 = 1$, so the 4th statement is incorrect.
Now, for the 1st statement, if it is true then
$$ \frac{f(x+y)-f(x)}{y} = 1 $$
If $x$ is seen as variable, and $y$ approach zero then we get
$$f'(x)=1$$, which means that $f(x) = x$.
My argument is that the 1st statement is also not true (perhaps inconclusive is more appropriate), lack of information.
 A: The first statement can be true or false, depending on $f$. The other three statements are always false.
For the first statement, $f(x)=x$ gives a supporting example and $f(x)=\lceil x\rceil$ gives a counterexample.
For the second and the fourth statements, your reasoning isn't correct. That $f(0+1)=1\le f(0)+1$ only implies that $0\le f(0)$. You cannot infer that $f(0)=0$ from it. However, one can infer that $f(x+1)=f(x)+1$ for every $x$ using squeezing principle:
$$f(x+5)\le f(x+4)+1\le f(x+3)+2\le f(x+2)+3\le f(x+1)+4\le f(x)+5\le f(x+5).$$
It follows from $f(1)=1$ that $f(n)=n$ for every integer $n$, and this falsifies statements 2 and 4.
The third statement is false, but again your reasoning is incorrect. You are correct in pointing out that $g(x)<f(x)$ when $1-x<0$, but this is not a counterexample to the statement that $g(x)\le f(x)\ \forall x\in\mathbb R$. To disprove the statement, you need to show that $g(x)\color{red}{>}f(x)$ for some $x$, and this is the case when $1-x\color{red}{>}0$.
