$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$ 
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.


In this case, since the equation is linear and given the "clue" that the coefficient of $x^2$ is non-zero, then consider $f(x) = a + b x + c x^2$. This yields the possible polynomial solutions, but not necessarily all possible solutions.
\begin{align}
x^2 &= (a + b x + c x^2) - 2 \, \left( a + \frac{b x}{2} + \frac{c x^2}{4} \right) +  \left( a + \frac{b x}{4} + \frac{c x^2}{16} \right) \\
&= \frac{b x}{4} + \frac{ 9 \, c \, x^2}{16}.
\end{align}
This leads to $b=0$ and $c = 16/9$ and the general solution of the form
$$f(x) = a + \frac{16 \, x^2}{9}.$$
