# Find maxima of the following expression using calculus (2x-sin2x)/x^2 ?

Aug 29, 2016

$x = - \frac{\pi}{2}$ minimum point
$x = \frac{\pi}{2}$ maximum point

#### Explanation:

Given $f \left(x\right) = \frac{2 x - S \in \left(2 x\right)}{x} ^ 2$

the stationary points are determined solving for $x$ the condition

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{4 \cos x \left(\sin x - x \cos x\right)}{x} ^ 3 = 0$

or

$\left\{\begin{matrix}\cos x = 0 \\ \tan x - x = 0\end{matrix}\right.$

with solutions

$x = 0 \cup \pm \frac{\pi}{2} + 2 k \pi , k = 0 , 1 , 2 , 3 , \cdots$

Here $x = 0$ is the main solution of $\tan x = x$

Now testing in the range $- 2 \pi \le x \le 2 \pi$ in

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(x\right) = \frac{2 \left(2 x + 4 x C o s \left(2 x\right) - 3 S \in \left(2 x\right) + 2 {x}^{2} S \in \left(2 x\right)\right)}{x} ^ 4$

we will obtain

${\lim}_{x \to 0} {d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(x\right) = 0$ inflexion point

and for $k = 0$

${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(- \frac{\pi}{2}\right) = \frac{32}{\pi} ^ 3$ minimum point
${d}^{2} / \left({\mathrm{dx}}^{2}\right) f \left(\frac{\pi}{2}\right) = - \frac{32}{\pi} ^ 3$ maximum point

there are infinite minima and maxima for $k = 1 , 2 , \cdots$ but their range is contained inside the former two for $k = 0$.

Note:

We have obviate the limit determination for $x \to 0$ in all the process steps.

For qualification justification see
https://socratic.org/questions/if-f-x-x-8-9-x-4-7-x-3-7-what-are-the-local-minima-and-maxima-of-f-x#300548

Attached the plot of $f \left(x\right)$ for $- 2 \pi \le x \le 2 \pi$