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.


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.

  • $\begingroup$ 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

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

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


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

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, $f(0) = 1$.

  • $\begingroup$ 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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