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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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    @J.M. If you're sending it to [math.se], please delete my answer before you do so.2012-07-08

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$0$ 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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    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.2012-07-08
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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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    You are right, I was being a bit quick and sloppy.2012-07-08
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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 $k=1$).