# Find $f(2)$ if $f$ satisfies $2f(x)-3f(\frac1x)=x^2$

The following expression is given, and we are asked to find $f(2)$. $$2f(x)-3f\left(\frac{1}{x}\right) =x^2$$

Does a unique and well defined answer exist? Why? and what is it?

We have that $f(1)=f(-1)=-1$.

• Note that we have no extra information, such as whether it is continuous or analytic . – Student tea Feb 20 '15 at 15:42

Make $x = 2$ and get $2f(2) - 3f(0.5) = 4$. Make $x = 0.5$ and get $2f(0.5) - 3f(2) = 0.25$. Solve the system: $$\begin{cases} 2f(2) - 3f(0.5) = 4 \\ -3f(2) + 2f(0.5) = 0.25\end{cases}.$$

• Nice. Same approach helps express $f$ explicitly. – Macavity Feb 20 '15 at 15:51

Hint: Note that $$\begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix} \begin{bmatrix} f(x)\\ f(1/x) \end{bmatrix} = \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix}$$ and $$\det\begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix}=-5\ne0$$

Now that over a year has passed, I think it is appropriate to finish the answer. \begin{align} \begin{bmatrix} f(x)\\ f(1/x) \end{bmatrix} &= \begin{bmatrix} 2&-3\\ -3&2 \end{bmatrix}^{-1} \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix}\\ &= -\frac15\begin{bmatrix} 2&3\\ 3&2 \end{bmatrix} \begin{bmatrix} x^2\\ 1/x^2 \end{bmatrix}\\ &=-\frac15\begin{bmatrix} 2x^2+3/x^2\\ 3x^2+2/x^2 \end{bmatrix}\\ \end{align} Therefore, $$f(x)=-\frac{2x^4+3}{5x^2}$$
• To the downvoter: assuming that this is not a downvote out of spite and since my hint leads directly to a formula for $f$, I would appreciate knowing what is wrong with this answer. – robjohn Jul 10 '17 at 22:14
For a general equation of the type $af(x)+bf(1/x)=g(x)$, for some known $g$ with $x\ne 0$, the way to solve is to consider the two equations $$\begin{pmatrix} a & b\\ b & a\\ \end{pmatrix}\begin{pmatrix} f(x)\\ f(1/x) \end{pmatrix}=\begin{pmatrix} g(x)\\ g(1/x) \end{pmatrix}$$ This will give you a solution for $f(x)$ as long as $a\ne b$. If $a=b$, then the equation holds only for $\{x:g(x)=g(1/x)\}$