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I am trying to calculate the maximum and minimum points (between $-3\pi$ and $3\pi$) of $$f(x)=\frac{\sin(x)}{x}$$
I have found the derivative of the function and let it equal to zero.
$$f'(x)=\frac{x\cos(x) - \sin(x)}{x^2}$$
$$f'(x)=0$$ $$\frac{x\cos(x) - \sin(x)}{x^2}=0$$ $$x\cos(x) - \sin(x)=0$$ $$x\cos(x)=\sin(x)$$ $$x=\tan(x)$$

I am unaware as to how to find $x$. I assume that once I find $x$, I can use a sign diagram or second derivative test to determine the minimum and maximum values. Any help would be highly appreciated.

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Equations which contains polynomial and trigonometric functions do not show explicit solutions and numerical methods should be used to find the roots.

The simplest root finding method is Newton; starting from a reasonable guess $x_0$, the method will update it according to $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)}$$ So, in your case, $$F(x)=x \cos(x)-\sin(x)$$ I think it is better to let it under this form because of the discontinuities of $\tan(x)$.

You can notice that if $x=a$ is a root, $x=-a$ will be another root. So, let us just focus on $0\leq x \leq 3\pi$. If you plot the function, you notice that, beside the trivial $x=0$, there are two roots located close to $5$ and $8$. These would be the guesses.

Using $F'(x)=-x \sin (x)$, the iterative scheme then write $$x_{n+1}=x_n-\frac{1}{x_n}+\cot (x_n)$$ Let us start with $x_0=5$; the method then produces the following iterates : $4.50419$, $4.49343$, $4.49341$ which is the solution for six significant figures.

I let you doing the work for the other solution.

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  • $\begingroup$ Thank you for your answer. After reading about Newton's method, I think I have been able to form a (mostly) correct answer for what I need. $\endgroup$ – Lint May 9 '15 at 4:02
  • $\begingroup$ You are very welcome ! I am glad to know that you learnt about Newton method. It is very simple and efficient. Cheers :-) $\endgroup$ – Claude Leibovici May 9 '15 at 4:08
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You don't need to solve the equation for $x$ to find the extreme point. You know that extremes happen when $x = \tan(x)$ and, therefore, taking the inverse $\tan$ on both sides, when $\mathrm{atan}(x) = x$. I will write $x_e$ to note the points where $f(x)$ has an extreme. Then

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin[\mathrm{atan}(x_e)]}{x_e}$.

But

$\displaystyle \sin[\mathrm{atan}(x_e)] = \frac{x_e}{\sqrt{1 + x_e^2}}$,

so, for each $x_e$ that is the point of an extreme, the value of the function at a point is

$\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$.

An alternative would be

$\displaystyle f(x_e) = \frac{\sin(x_e)}{x_e} = \frac{\sin(x_e)}{\tan(x_e)} = \cos(x_e)$, which is actually the same result, since, using again that $x_e = \mathrm{atan}(x_e)$, we have

$\displaystyle \cos[\mathrm{atan}(x_e)] = \frac{1}{\sqrt{1 + x_e^2}}$.

In any case, you see that $\displaystyle f(x_e) = \frac{1}{\sqrt{1 + x_e^2}}$ is maximum at $x_e = 0$, as it decreases everywhere else. At this point, $f(0) = 1$.

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  • $\begingroup$ There are infinitely many solutions to $\tan x = x$, but by construction the only solutions among these that are also solutions of $x = \arctan x$ are those with $|x| < \frac{\pi}{2]$; in fact, by differentiating one can show there is only one solution to the latter, namely, $x = 0$. $\endgroup$ – Travis Willse May 8 '15 at 9:49

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