Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that: $\;x^2\cdot f(x)+f(1-x)=2x-x^4,\;\forall\; x\in \mathbb R.$
My solution:
Replace $x$ by $(1-x)$ and by eliminating $f(1-x)$.
I obtained $f(x)=1-x^2$ as provided in my book.
My doubt:
How to check that there is no other function satisfying Above property?
Also, I think $f(x)=0,\;\forall\;x\in \mathbb R$ can be solution too.
Is my thinking correct?
Best Answer
Your solution is incomplete, because the solution is not required to be continuous. Your system of equations $$ \begin{pmatrix} x^2 & 1 \\ 1 & (1-x)^2 \end{pmatrix} \begin{pmatrix} f(x) \\ f(1-x) \end{pmatrix} = \begin{pmatrix} 2x-x^4 \\ 2(1-x) - (1-x)^4 \end{pmatrix} $$ has infinitely many solutions if $x=\varphi =\frac{1+\sqrt{5}}{2}$ or $x=1-\varphi =\frac{1-\sqrt{5}}{2}$. In this case, the matrix is not invertible, and we get $$ \begin{pmatrix} f(\varphi) \\ f(1-\varphi) \end{pmatrix} =\begin{pmatrix} -\varphi \\ \varphi-1 \end{pmatrix} +\lambda \begin{pmatrix} -1 \\ 1+\varphi \end{pmatrix} $$ So for each $\lambda \in \mathbb{R}$, the following is a valid solution: $$ f(x)= \begin{cases} 1-x^2 & \text{ for } x\in\mathbb{R}\setminus\{\varphi, 1-\varphi\} \\ -\varphi -\lambda & \text{ for } x=\varphi \\ \varphi-1 +\lambda(1+\varphi) & \text{ for } x=1-\varphi \end{cases} $$