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let's say we have the function $$y = \frac{x}{x^2+1}$$

we see that y': $$y' =\frac{-x^2+1}{(x^2+1)^2}$$

by the second derivative test, we see that the points $x=1$ is a local maximum and $x=-1$ is a local minimum that's the only clue I have about the symmetry of this graph.

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    $\begingroup$ What kind of symmetries? The function is odd, so it's symmetric about the origin. Also, for $x \ne 0$ you can write it as $y = \frac{1}{x + 1/x}$ so it's invariant under $x \mapsto \frac{1}{x}$. $\endgroup$
    – dxiv
    Jun 6 at 21:34
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    $\begingroup$ You detect whether a function has symmetries by specifically testing for said symetries. What kind of symmetries do you know, how are they defined, and which ones do you suspect apply? $\endgroup$
    – Arthur
    Jun 6 at 21:35
  • $\begingroup$ There are many ways to see a function is symmetric, for example you could note that the Fourier transform comes out purely imaginary. But this is really overcomplicating things... indeed one might more sensibly go the other direction, i.e. use the fact that the function is symmetric as an aid for computing the Fourier transform. $\endgroup$
    – leftaroundabout
    Jun 7 at 13:57

2 Answers 2

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We can check whether the function is even or odd:

  • If function is odd, meaning $f(-x)=-f(x)$ than it is symmetrical with respect to the origin
  • If function is even, meaning $f(-x)=f(x)$ than it is symmetrical with respect to $y$-axis

In this example, we can see that $$f(-x)=\frac{-x}{x^2+1}=-f(x)$$ thus this function is odd, meaning it is symmetrical with respect to the origin.

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Notice that

\begin{align*} f(x) = \frac{x}{x^{2} + 1} \Rightarrow f(-x) = -\frac{x}{(-x)^{2} + 1} = -\frac{x}{x^{2} + 1} = -f(x) \end{align*}

hence we conclude its graph is symmetric about the origin.

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