$f\left(\frac{x+y}{3}\right)=\frac{2+f(x)+f(y)}{3}$
find the range of $f(|x|)$ , please tell me the solution and weather we can do it using the limits : first principle. Im only a 12th grader so dont use any college level formulas(if they exist for this kind of thing)
Let us find differentiable function $f\colon\Bbb R\to\Bbb R$ such that
$$ f\left( \frac{x+y}3 \right) = \frac{2+f(x)+f(y)}{3},\ f'(2) = 2.$$
First substitute $x \mapsto 3x$ and $y \mapsto 3y$ to get
$$3f(x+y) = 2 + f(3x)+f(3y)\tag{1}$$
Then, for $x = y = 0$ we get $3f(0) = 2 + f(0) + f(0)$ which implies that $f(0) = 2$.
Letting $y = 0$ gives $3f(x) =2+f(3x) +f(0)$ which implies that
$$f(3x) = 3f(x)-4\tag{2}$$
Now, use $(2)$ in $(1)$ to get $3f(x+y)= 2 + (3f(x)-4)+(3f(y)-4)$, i.e.
$$f(x+y) = f(x)+f(y)-2\tag{3}$$
Define $g(x) = f(x)-2$. If we subtract $2$ from both sides of equation $(3)$, we get $$g(x+y) = g(x)+g(y)\tag{4}$$
Equation $(4)$ is known as Cauchy functional equation and I strongly recommend that you read about it at this beautiful MSE faq.
Since $f$ is differentiable, so is $g$. Therefore, $g(x) = ax$, for some constant $a\in\Bbb R$ (you can find why if you follow the link). Then, $f(x) = ax + 2$.
Finally, $f'(x) = a$ and condition $f'(2) = 2$ then implies that $a = 2$. Thus, $$f(x) = 2x+2$$
Hopefully, you can now finish the exercise yourself.
If you want to avoid mentioning Cauchy functional equation, you could use differentiability of $f$ more directly.
\begin{align} f'(x) &= \lim_{y\to 0}\frac{f(x+y)-f(x)}{y}\\ &\stackrel{(3)}{=} \lim_{y\to 0}\frac{f(x)+f(y)-2-f(x)}{y}\\ &= \lim_{y\to 0}\frac{f(y)-2}{y} \tag{5}\\ \end{align}
In particular, $f'$ is constant, and thus $f(x) = ax+b$.
Returning to $(5)$ and using $f'(2) = 2$, we get
$$2 = \lim_{y\to 0}\frac{ay + b - 2}{y} = a + \lim_{y\to 0}\frac{b - 2}{y}$$
Now, $\lim_{y\to 0}\frac{b - 2}{y} = \pm\infty$, unless $b = 2$. It follows that $a = 2$.
