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I am given the function $f(x)= x/(x^2-1)$. I find that the derivative is:

\begin{equation} \frac{x^2+1}{(x^2-1)^2} \end{equation}

So, I note that $f'(x)$ is undefined at $x = \pm 1$. However, those points also are NOT in the domain of the original function.

So, my question is, since there are no critical points, is there anything I can conclude from $f'(x)$?

I did go on to find $f''(x) =2x(x^2 +3)/(x^2 +1)^3$ and I can see that this second derivative is zero (inflection point) at $x=0$.

The text says I should look on either side of $x=-1$ and $x=1$ to evaluate concavity. However, it would seem to me that $x=1$ and $x=-1$ are NOT critical points so what causes me to need to look at those points?

Thanks

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    $\begingroup$ Critical points (or rather, maxima and minima) occur at points where either $f'(x)=0$ or $f'(x)$ is undefined, The same goes for concavity, so you still have to check those specific cases. $\endgroup$ – Edward Jiang Apr 26 '16 at 19:32
  • $\begingroup$ QYou should look *on either side of * $x=\pm1$, i.e., plug in $x=-2, 0, 2$ in $f'$ an determine its sign to find out that $f$ is strictly decreasing an each of the intervals $]-\infty,-1[$, $]-1,1[$, and $]1, \infty[$. $\endgroup$ – Michael Hoppe Apr 26 '16 at 19:35
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    $\begingroup$ I thought it was a critical point ONLY if the cp is IN the domain of the f(x). $\endgroup$ – user163862 Apr 26 '16 at 19:44
  • $\begingroup$ Maybe the way I should think about it is there can be maxima or minima near the critical points even if those points are not defined in f(x) -- for example if they are asymptotic at those points. $\endgroup$ – user163862 Apr 26 '16 at 19:46
  • $\begingroup$ *The text says I should look on either side of x=-1 and x=1 to evaluate concavity. * so there are no maxima or minima but quite sure a change in concavity. $\endgroup$ – N74 Apr 26 '16 at 19:53
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As you say, there are no critical points, so you can conclude that f has no relative extrema.

Furthermore, since $\displaystyle f^{\prime}(x)=-\frac{x^2+1}{(x^2-1)^2}<0$ for $x\ne\pm1$, you can also conclude that

$\hspace{.4 in}f$ is decreasing on the intervals $(-\infty, -1),\;(-1,1),\;\text{and }(1,\infty)$.


Since the second derivative $f^{\prime\prime}$ equals 0 at $x=0$ and is undefined for $x=\pm1$,

you have to determine the sign of $f^{\prime\prime}$ on each of the intervals $(-\infty,-1), (-1,0), (0,1), \text{and }(1,\infty)$

to determine the concavity of the graph.

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