What $x$ makes $\frac{x}{(a^2 + x^2)}$ maximum? The problem (Calculus Made Easy, Exercises IX, problem 2 (page 130)) is:

What value of $x$ will make $y$ a maximum in the equation
$$y = \frac{x}{(a^2 + x^2)}$$

I successfully differentiate, equate to zero, and wind up with

$$x^2 = a^2$$

Which gives me the answer of

$$x = a$$

This is correct.  But why isn't $x = -a$ also correct?

We are given
$$y = \frac{x}{a^2 + x^2}$$
where $a$ is a constant.  
Differentiating with respect to $x$ using the Quotient Rule yields
\begin{align*}
y' & = \frac{1(a^2 + x^2) - x(2x)}{(a^2 + x^2)^2}\\
   & = \frac{a^2 + x^2 - 2x^2}{(a^2 + x^2)^2}\\
   & = \frac{a^2 - x^2}{(a^2 + x^2)^2}
\end{align*}
Setting the derivative equal to zero yields the critical points $x = \pm a$.  
We can apply the First Derivative Test.
First Derivative Test.  Assume $f$ is continuous on a closed interval $[u, v]$ and $f$ is differentiable everywhere in the open inteval $(u, v)$ except possibly at $c$.  
(a) If $f'(x) > 0$ for all $x < c$ and $f'(x) < 0$ for all $x > c$, then $f$ has a relative maximum at $x = c$.
(b) If $f'(x) < 0$ for all $x < c$ and $f'(x) > 0$ for all $x < c$, then $f$ has a relative minimum at $x = c$.
If $a = 0$, then $y = \dfrac{1}{x} \implies y' = -\dfrac{1}{x^2}$, so the function has no critical points and no relative extrema.
Assume $a \neq 0$.  Since it has not been specified whether $a > 0$ or $a < 0$, the critical points occur at $x = -|a|$ and $x = |a|$.  If we perform a line analysis on the derivative, we see that $y'$ changes from negative to positive at the critical point $x = -|a|$ and from positive to negative at the critical point $x = |a|$.  

Thus, by the First Derivative Test, the function has a relative maximum at $x = |a|$ and a relative minimum at $x = -|a|$.  If it is specified that $a > 0$, you can replace $|a|$ by $a$.
