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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?

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    $\begingroup$ $a^2+x^2$ 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. $\endgroup$ – Ak19 Jul 20 at 0:52
  • $\begingroup$ 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? $\endgroup$ – Mike Jul 20 at 1:02
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    $\begingroup$ 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 $\endgroup$ – Ak19 Jul 20 at 1:05
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Because $\dfrac{-a}{a^2+(-a)^2}$ is a minimum for $a>0$

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

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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 $x=-a$.

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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.

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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 $x = |a|$.

line_analysis_of_derivative

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 $a$.

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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 $y = x^3$.

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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.

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