Functional equation

Calculus Level 3

Determine the number of differentiable real-valued functions $f$ that satisfy $f^{ 2 }\left( x \right) +2{ x }^{ 2 }=3xf\left( x \right)$ for all $x\in\mathbb{R}$ .

The answer is 2.

1 solution

Félix de Montemar
Mar 24, 2016

This equation is equivalent to $f^2\left(x\right)-3xf\left(x\right)+2x^2=0$ and then, $\left( f\left( x \right) -x \right)\left( f\left( x \right) -2x \right)=0$ for all $x\in\mathbb{R}$ , so pointwise $f\left( x \right)=x$ or $f\left( x \right)=2x$ . Since $f$ is a continuous function, then there are 4 solutions.

$f\left( x \right)=x$
$f\left( x \right)=2x$
$f\left( x \right) =\begin{cases} x\quad x\ge0 \\ 2x\quad x<0 \end{cases}$
$f\left( x \right) =\begin{cases} 2x\quad x\ge0 \\ x\quad x<0 \end{cases}$

Moreover, as $f$ is a differentiable function, $f$ can not be the third and fourth solutions as they are not differentiable at $x=0$ , so $f(x)=x$ and $f(x)=2x$ are the two solutions

Moderator note:

Your solution is still incomplete. What you have shown, is that pointwise we must have $f(x) = x$ or $f(x) = 2x$ .

You have not shown yet that we must have the function be everywhere equal to $f(x) = x$ or $f(x) = 2x$ .

Your solution is still incomplete. What you have shown, is that pointwise we must have $f(x) = x$ or $f(x) = 2x$ .

You have not shown yet that we must have the function be everywhere equal to $f(x) = x$ or $f(x) = 2x$ .

For example, if we relaxed the condition to continuous functions, then there are 4 solutions. The other 2 are of the form

$f(x) = \begin{cases} x & x \geq 0 \\ 2x & x < 0 \\ \end{cases}$

Calvin Lin Staff - 5 years, 2 months ago

But, in this case I think that the function is not differentiable at x=0. I did not change my first answer, but I wrote a little explanation in a comment

Félix de Montemar - 5 years, 2 months ago

I'm pointing out that it's a very common mistake to take a pointwise property and assume that it holds globally, which is what you have done.

Please edit the solution directly so that the conclusion is completely justified. I don't think it's fair to expect others to root through all of the comments, especially if you didn't indicate to do so.

Calvin Lin Staff - 5 years, 2 months ago

A quadratic question that looks like a functional equation...
I think that continuity should be assumed, otherwise some functions like $f(x)=x$ somewhere, $f(x)=2x$ elsewhere also satisfy the condition given.

展豪張 - 5 years, 2 months ago

You are right, I just correct it. My apologies!

Since $f$ is differentiable, the only two options are $f(x)=x$ and $f(x)=2x$ . In any other case, $f$ would have a discontinuity or, if $f$ was continuous, $f$ would not be differenciable at $x = 0$