# 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?

• $a^2+x^2$ is always positive. We can clearly see that if $a>0$ and $x=a$ , $$y = \frac{a}{2a^2} = \frac{1}{2a} > 0 \text{ maximum }$$. But if $x = -a$ , $y = -\frac{1}{2a} < 0$. So it all depends on $a$. If we assume $a$ to be positive only $x=a$ will lead to maximum as the other term corresponding to $x=-a$ is negative. – Ak19 Jul 20 at 0:52
• But $y$ being greater than zero doesn't make it a maximum, does it? I mean, there are functions where the maximum value for $y$ is negative. Or are you saying that $\frac{1}{2a}$ is greater than $\frac{-1}{2a}$, therefore the former has to be the maximum of the two? – Mike Jul 20 at 1:02
• As you already know that there's a maximum corresponding to $x=a$ I just added to show that $y=- 1/(2a)$ is negative and minimum – Ak19 Jul 20 at 1:05

Because $$\dfrac{-a}{a^2+(-a)^2}$$ is a minimum for $$a>0$$

Strictly speaking, the maximum is at $$x=\lvert a\rvert$$.

Substitute $$x=\pm a$$ in the original expression for $$y$$ and compare. If $$a>0$$, then yes, $$x=a$$ is the value that make $$y$$ maximum, if $$a<0$$ then is $$x=-a$$.

When you differentiated and got $$x^2=a^2$$ for a turning point.Now we know that turning points are $$x=a$$ and $$x=-a$$. To check whether it is a maximum or a minimum, $$\frac{d^2y}{dx^2}$$ must be calculated.If $$\frac{d^2y}{dx^2}$$ is $$<$$ $$0$$ for $$x$$ then $$f(x)$$ is maximum ,else if $$\frac{d^2y}{dx^2}$$ is $$>$$ $$0$$ for $$x$$ then, $$f(x)$$ is minimum.

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 $$a$$.

It is a necessary but not sufficient condition for the maximum of a function on $$\mathbb{R}$$ to satisfy $$y' = 0.$$

That is

$$y(x) \text{ is a maximum} ~~~ \text{implies} ~~~ y'(x) = 0$$

But

$$y'(x) = 0 ~~~\text{does NOT imply} ~~~ y(x) \text{ is a maximum}$$

As you have seen in your own example. You may also like to consider a function such as $$y = x^3$$.

Using AM-GM: $$a^2+b^2\ge 2ab$$ or $$a^2+b^2\ge 2|a||b|$$.

For $$x\ne 0$$: $$\frac{x}{a^2 + x^2}\le \frac{|x|}{a^2+x^2}\le \frac{|x|}{2|a||x|}=\frac1{2|a|},$$ the equality occurs for $$x=a>0$$ or $$x=-a>0$$.

If $$a=0$$, then $$y=\frac1x$$ does not have a maximum.