$f(x)$ is given as $$\frac{e^x}{1+x^2}$$ and $g(x)=f'(x)$ then $g(x)$ has how many points of local maxima and local minima?

Is $x=1$ a point of local minima for $g(x)$ and is there a point of local maxima for $g(x)$ in the interval $(-1,0)$?

How to answer these questions without going through too much calculations like manually finding out $g'(x)$?Can this question be solved quickly by any graphical approach?

This appeared on a Multiple Choice Type Question in my test today and I'm still wondering how the paper setters expected us to solve it in 2-3 minutes.So any shortcut you can think of ?

  • $\begingroup$ What do you mean by "graphical approach"? Do you mean using a grapher? Most graphers can directly graph a derivative, so just graph that and look at the extrema. $\endgroup$ – Rory Daulton Apr 17 '16 at 18:44
  • $\begingroup$ @RoryDaulton no.I mean manually.No grapher allowed.By graphically I meant plotting the graph approximately by hand(on pen and paper). $\endgroup$ – user220382 Apr 17 '16 at 18:45
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    $\begingroup$ It takes less than 2 mins to get $g(x)=e^x\frac{(1-x)^2}{(1+x^2)^2}$. It is obvious that this has a zero and a minimum at $x=1$. It should also be clear that it tends to 0 as $x\to-\infty$, whereas it is 1 at $x=0$, so it has a max for some negative $x$. I agree that pinning the max down to $(-1,0)$ will probably take you over 3 mins! $\endgroup$ – almagest Apr 17 '16 at 21:29

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