How to find the maximum and minimum of $\dfrac{\sin x}{x^2+1}$? How can we find the values of $x$ that give the maximum and minimum of $$\frac{\sin{x}}{x^2+1}$$
I took a lucky guess and found that $\dfrac\pi4$ was fairly close to giving the max, but how does one do this using calculus?
 A: As @pew points out, the critical point in the first quadrant is not easy to obtain analytically, but this is where the maximum occurs.  The function:
$$ y = \frac{\sin x}{x^2 + 1} $$
is defined on the entire real line.  Since it is an odd function, its derivative:
$$ y' = \frac{(x^2+1)\cos x - 2x \sin x}{(x^2 + 1)^2} $$
is even, and the minimum is attained at the reflection of the maximum in the origin.  Note that the derivative is negative in the second quadrant (where cosine is negative and sine is positive), so the function $y(x)$ is decreasing there.
Thus it suffices to find the critical point in the first quadrant, since the oscillations of the function $y(x)$ are "dampening" as $x$ increases.
Setting $y'=0$ as in the First Derivative Test gets us this transcendental equation:
$$ (x^2 + 1)\cos x - 2x \sin x = 0 $$
We could of course use "calculus" to find a numerical root in $[0,\pi/2]$ using Newton iterations.  However I suspect it will be a nice challenge to rewrite the critical point equation above in a form $x = g(x)$ so that fixed point iteration on $g(x)$ starting at $x = \pi/4$ converges linearly (by contraction).

 $$ x = \tan^{-1}\left(\frac{x^2 + 1}{2x}\right) $$

