Question

F(x) is continuous, given that f(x+5) >= f(x) + 5 and f(x+1)<= f(x) + 1. If g(x) = f(x) +1 - x, then g(2025)??

Answer 1

1

Answer 2

Solution:

1. The given inequalities are

   $$f(x+5) \geq f(x) + 5\quad \text{and}\quad f(x+1) \leq f(x) + 1.$$

2. For any integer $n$ we can write

   $$f(x+n) = f(x) + [f(x+n)-f(x)].$$

   In particular, by repeatedly applying the second inequality,

   $$f(x+5) \leq f(x) + 5.$$

3. Thus, we have

   $$f(x+5) \geq f(x) + 5\quad \text{and}\quad f(x+5) \leq f(x) + 5,$$

   so

   $$f(x+5) = f(x) + 5.$$

4. In the same way, one can show that

   $$f(x+1) = f(x) + 1,$$

   for all $x$. Define

   $$h(x) = f(x) - x.$$

   Then,

   $$h(x+1) = f(x+1) - (x+1) = [f(x)+1] - (x+1) = f(x) - x = h(x),$$

   so $h(x)$ is 1-periodic.

5. Now, define

   $$g(x) = f(x) + 1 - x.$$

   Expressing $f(x)$ in terms of $h(x)$,

   $$f(x) = x + h(x) \quad\Rightarrow\quad g(x) = x + h(x) +1 - x = h(x) + 1.$$

   Since $h(x)$ is 1‑periodic, $g(x)$ is 1‑periodic.

6. Therefore, for an integer $n$ (such as 2025),

   $$g(2025)= h(2025) + 1 = h(0) + 1.$$

7. Without additional initial conditions, the general solution is $f(x)= x+ c$ (where $c=h(0)$ is constant). For the “minimal” or standard choice $c=0$ (i.e. $f(x)=x$), we get

   $$g(2025)=0+1=1.$$

Explanation (minimal):

• From $f(x+1) \leq f(x) + 1$ applied repeatedly, we deduce $f(x+1)=f(x)+1$.

• This implies $f(x)=x+h(x)$ with $h(x)$ 1‑periodic.

• Consequently, $g(x)=f(x)+1-x=h(x)+1$ is also 1‑periodic.

• In particular, $g(2025)=h(0)+1$. Choosing $h(0)=0$ (i.e. $f(x)= x$) yields $g(2025)= 1$.