Find the function that matches the equation

[YoungPhysicist]

Homework Statement

$3f(x)+2f(\frac{1}{x}) = x$, solve $f(x)$

Homework Equations

Not sure.Maybe the ones of inverse functions.

The Attempt at a Solution

The only thing that I came up so far is that the function’s highest order term is $x$ because if there are higher orders,it will show up at the right side of the equation.

Then I am stuck. The question is after inverse functions, so maybe it has something to do with that.

 

[DrClaude]

Think about it. When you have $f(1/x)$, any resulting term not in the form of $x$ must cancel out with a term that comes from $f(x)$.

 

[RPinPA]

If you're guessing that $f(x)$ is of the form $f(x) = ax + b$ then $f(1/x) = (a/x) + b$. Plug those into the equation, equate coefficients of each power of $x$ and see where that gets you. I don't think it will get you very far, I think you'll find it's a contradiction. But I highly recommend doing it anyway to get used to the algebra of how you're going to attack this.

Now try a different form of $f(x)$, and do the same thing. Write down the expression for $f(1/x)$. Plug them in. Solve for free parameters. See if you get a solution.

@DrClaude has given you a big hint. The reason that form above won't work is that there's nothing to cancel out the $a/x$ term on the left. So what can you do about that?

 

[SammyS]

Homework Statement

$3f(x)+2f(\frac{1}{x}) = x$, solve $f(x)$

Homework Equations

Not sure.Maybe the ones of inverse functions.

The Attempt at a Solution

The only thing that I came up so far is that the function’s highest order term is $x$ because if there are higher orders,it will show up at the right side of the equation.

Then I am stuck. The question is after inverse functions, so maybe it has something to do with that.

Substitute $1/x$ for $x$ and see what that gets you.

 

[YoungPhysicist]

OK. I think I got it.
Since the terms other than $x$ cancels out, and term $x$ reversed is $1/x$, so there must be a $1/x$ term.

$f(x) = ax + b/x$

When I substitute that back, I get: $3ax+\frac{3b}{x}+\frac{2a}{x}+2bx = x$
From that I can get the simultaneous equation:

$3a+2b-1 = 0$
$3b+2a = 0$

The solution of that is $a = 3/5, b = -2/5$

So the function is:

$$f(x) = \frac{3}{5}x - \frac{2}{5x}$$

I put that in geogebra, and looks like it is correct.Thanks everyone!
Edit: Oh, and $x \ne 0$.

 

[RPinPA]

Yep, that's what I was thinking. But @SammyS had an elegant suggestion which doesn't require you to make an initial guess on the form of $f(x)$.

For any nonzero $x$, $3f(x) + 2f(\frac 1 x) = x$.
But by evaluating at $\frac 1 x$ you must also have $3f(\frac 1 x) + 2f(x) = \frac 1 x$.
That gives you simultaneous equations for $f(x)$ and $f(\frac 1 x)$.

Eliminating $f(\frac 1 x)$ by multiplying by appropriate constants and subtracting:
$3 \left [3 f(x) + 2 f(\frac 1 x) \right ] - 2\left [ 3f(\frac 1 x) + 2 f(x) \right ] = 3x - 2\left (\frac 1 x \right )$
$9f(x) - 4 f(x) = 5 f(x) = 3x - \frac 2 x$
$f(x) = \frac {3x} 5 - \frac 2 {5x}$

 

[YoungPhysicist]

Eliminating $f(\frac 1 x)$ by multiplying by appropriate constants and subtracting:

Ah,Sorry @RPinPA , I don't quite get this part.

 

[Mark44]

Ah,Sorry @RPinPA , I don't quite get this part.

@RPinPA is solving a system of two equations in two unknowns, with the unknowns being f(x) and f(1/x).

If you have
ax + by = c and
dx + ey = f

you can eliminate the y terms by multiplying the first equation by -e and the second equation by b, and then adding the two equations.

This gives you
-aex - bey = -ce
bcx + bey = bf

If you add these equations, you get a single equation in x only.

 

[YoungPhysicist]

@RPinPA is solving a system of two equations in two unknowns, with the unknowns being f(x) and f(1/x).

If you have
ax + by = c and
dx + ey = f

you can eliminate the y terms by multiplying the first equation by -e and the second equation by b, and then adding the two equations.

This gives you
-aex - bey = -ce
bcx + bey = bf

If you add these equations, you get a single equation in x only.

Oh.I see.Just not very used to simultaneous equation with functions being the variable.That kind of distracts me from treating it normally.

Thanks!

 

[RPinPA]

Ah,Sorry @RPinPA , I don't quite get this part.

Sorry, I meant "multiply the first equation by 3 and the second equation by 2 and then subtract".
I thought that was clear since that's exactly what I did on the very next line. And you can see that equation and see that $f(\frac 1 x)$ goes away.