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I was doing a calculus problem and got to this point where I neededed to obtain $f(x)$ for later integration. $f(x)$ is continuous and satisfies the equation $xf(x) = x^4 + 2x^2 + C,$ for some constant $C$ and for all $x \in \mathbb{R}$. Here's my argument:

Since the equation holds for all $x \in \mathbb{R},$ it must hold for $x = 0 \implies C = 0 \implies xf(x) = x^4 + 2x^2.$

If $x \neq 0,$ $f(x) = x^3 + 2x = g(x)$

For $f(x)$ to be continuous at $x = 0$, $f(0) = lim_{x\to0} f(x) = 0^3 + 2.0 = 0 = g(0)$

$\implies f(x) = g(x) = x^3 + 2x, \forall x \in \mathbb{R}.$

I'm not sure if I was overthinking here, but I found this argument fairly bulky. Please advise! Thank you.

ten_to_tenth
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    If $f$ is a piecewise function, $f = g$ whenever $x\neq0$ but $f(x) = $any constant when $x =0$. However, if all you need is to integrate f, the integral is unaffected by single, "removable" discontinuities. – nickalh Apr 13 '24 at 15:20
  • @nickalh Hi, if for one value of $x = 0,$ $f(x)$ could be any constant, doesn't that mean $f(x)$ is no longer a function? – ten_to_tenth Apr 13 '24 at 15:22
  • No, for any given definition of $f$, it can only take on a single value $f(0)=D$, Maybe call it a family of functions, $f_D(x) = g$ except $f_D(0) = D$, for $D \in \mathbb(R)$ – nickalh Apr 13 '24 at 15:30
  • @nickalh Since $f(x)$ is stated to be continuous, doesn't it mean only one such instance of $f(x)$ in which $D = 0$ satisfies this condition? – ten_to_tenth Apr 13 '24 at 15:34
  • Hmmm, in the form stated in your question, D has no effect on f being continuous. $0\cdot D = 0$. Beyond that, I need to move on to other things, sorry. – nickalh Apr 13 '24 at 15:37
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    We're quibbling over nonessential details. Any single point of a function has no effect on the integral. I'm not sure if this will help you or confuse you, but google "integral of Thomae function". If you're in Calculus I, I recommend simply accepting that any single point has no effect on the integral. An intuitive, but not thorough way to understand this follows. The point, $(0,D)$ is a line segment, with zero width. The "area" of this line segment is $A = 0\cdot D = 0$.

    https://math.stackexchange.com/questions/1104782/integral-of-thomaes-function

     – nickalh Apr 13 '24 at 16:19
  • You didn't specify, what level you are. In the future, please do so for every question you post. Calculus I, Calculus II, Calculus III, Undergraduate- Analysis I, Analysis II, Graduate Analysis I or II? It makes a huge difference as to which answers will make sense to you and what types of answers would be acceptable to your professor. – nickalh Apr 13 '24 at 16:21
  • Also, https://en.wikipedia.org/wiki/Thomae%27s_function Note: Thomae's function is sometimes called Dirichlet's function, but Dirichlet's function can also mean a related function, to which these properties do not apply. – nickalh Apr 13 '24 at 16:22
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    Your reasoning is perfect, $f$ has to be zero at zero by continuity! You did not overthink it. The argument of @nickalh about integrability is true but only if you wish to integrate (the value does not change if $f$ is different on one point) – julio_es_sui_glace Apr 13 '24 at 16:33
  • Thank you everyone! ;) – ten_to_tenth Apr 14 '24 at 01:09

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