# Finding the critical points of $\sin(x)/x$ and $\cosh(x^2)$

Could someone help me solve this:

What are all critical points of $f(x)=\sin(x)/x$ and $f(x)=\cosh(x^2)$?

Mathematica solutions are also accepted.

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## migrated from mathematica.stackexchange.comJul 8 '12 at 15:13

This question came from our site for users of Mathematica.

sounds like you should be asking this on Mathematics, although the last part technically makes it on-topic here. Also, since this is homework, why don't you start by telling us what you've tried and where you're getting stuck? –  R.M Jul 8 '12 at 14:48
What do you know about finding critical points? –  Guess who it is. Jul 8 '12 at 14:57
for sin(x)/x I got first derivative. then i calculated value of x, which is tan(x). now what to do further. –  pr3dat0r Jul 8 '12 at 14:57
as now I have to put value of x in second derivative further. –  pr3dat0r Jul 8 '12 at 15:02
$\tan\,x$ is not the correct derivative for $\frac{\sin\,x}{x}$... –  Guess who it is. Jul 8 '12 at 15:05

$0$ is the only critical point of $x \mapsto \cosh x^2$ since its derivative $x\mapsto 2x\sinh x^2$ vanishes only at $x=0$. For $f(x)=\frac{\sin x}{x}$ first notice that $0$ is a removable singularity since $\lim_{x \to 0}f(x)=1$, so we can set $f(0)=1$. Then $$f'(x)=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x) \quad \forall x \ne 0$$ with $$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{1-\frac{x^2}{6}-1}{x}=\lim_{x \to 0}\frac{-x}{6}=0.$$ Thanks to the Intermediate Value Theorem one shows that for every positive integer $n$ the equation $x-\tan x=0$ possesses a unique solution $$x_n \in ((n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi).$$ By symmetry the set of critical points of $f$ is $\{0,\pm x_n: \ n \in \mathbb{N}\}$.

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Strictly speaking, he did not say the domain is the real numbers, so maybe complex critical points should be discussed (and dismissed). –  GEdgar Jul 8 '12 at 16:53
Which is it? discussed or dismissed? –  Mercy Jul 8 '12 at 16:54
I was contemplating including complex critical points in my answer, but decided it was above the OP's level more than likely as this seems like a traditional exercise from an intro calculus textbook. Also, is it just me or I can't click his profile (in case he included education info in there so we could see how to write answers more applicable for him perhaps)? –  Joe Jul 8 '12 at 16:55
Discussed or dismissed? Both, in that order. –  GEdgar Jul 8 '12 at 16:57
In fact, if $z \cos(z) - \sin(z) = 0$, $f(x) = \sin(z x)$ satisfies the differential equation $f'' = - z^2 f$ with boundary conditions $f(0)=0$ and $f'(1) - f(1) = 0$. It follows from the theory of Sturm-Liouville equations that $z^2$ must be real. The case where $z^2 < 0$ (i.e. $z$ purely imaginary) is easy to dismiss. –  Robert Israel Jul 8 '12 at 17:16

As mentioned, there is a solution in each interval $(k \pi, (k+1)\pi)$. This solution can't be expressed in "closed form", but there is a series in negative powers of $k$:

$$x = (k+1/2)\pi -{\frac {1}{k\pi }}+{\frac {1}{2 \pi \,{k}^{2}}}-{\frac {3\,{ \pi }^{2}+8}{12{\pi }^{3}{k}^{3}}}+{\frac {{\pi }^{2}+8}{8{\pi }^{3} {k}^{4}}}-{\frac {15\,{\pi }^{4}+240\,{\pi }^{2}+208 }{{240 \pi }^{5}{k}^{5}}}+{\frac {3\,{\pi }^{4}+80\,{\pi }^{2}+208}{96{\pi }^{5}{k}^{6}}}+\ldots$$

It looks to me like this converges for $k \ge 1$ (I'm not sure about $k=1$).

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Taken from here :

$x=c$ is a critical point of the function $f(x)$ if $f(c)$ exists and if one of the following are true:

1. $f'(c) = 0$
2. $f'(c)$ does not exist

The general strategy for finding critical points is to compute the first derivative of $f(x)$ with respect to $x$ and set that equal to zero.

$$f(x) = \frac{\sin x}{x}$$

Using the quotient rule, we have:

$$f'(x) = \frac{x\cdot \cos x - \sin x \cdot 1}{x^2}$$

$$f'(x) = \frac{x \cos x}{x^2} - \frac{\sin x}{x^2}$$

Dividing through by $x$ for the left terms, we now have:

$$f'(x) = \frac{\cos x}{x} - \frac{\sin x}{x^2}$$

Now set that equal to zero and solve for your critical points. Do the same for $f(x) = \cosh(x^2)$. Don't forget the chain rule!

For $f(x) = \cosh (x^2)$, recall that $\frac{d}{dx} \cosh (x) = \sinh (x)$. So,

$$f'(x) = \sinh(x^2) \cdot \frac{d}{dx} (x^2)$$

$$f'(x) = 2x \sinh(x^2)$$

$$0 = 2x \sinh(x^2)$$

$x = 0$ is your only critical point along the reals.

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There is a mistake at the penultimate line. "... because $\sinh (x^2) = 0$ when $x=0$" is not the logical implication you are looking for. –  D. Thomine Jul 8 '12 at 17:56
You are right, I was being a bit quick and sloppy. –  Joe Jul 8 '12 at 17:58