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I'm trying to find the minimum and maximum points of the following equation: $y = x \sin(\ln|x|)$ where $x > 0$ which, when graphed, looks something like this:

the graph of the equation

I tried deriving the equation using the chain rule:

$$\begin{align}\frac{dy}{dx} &= \frac{d}{dx}\left(x \cdot \sin(\ln|x|)) \right) \\ & = \sin(\ln|x|) + x \cos(\ln|x|) \frac{1}{x} \\ & = \sin(\ln|x|) + \cos(\ln|x|) \end{align}$$

...and setting it equal to zero:

$$0 = \sin(\ln|x|) + \cos(\ln|x|)$$

I set $\ln|x|$ equal to $\frac{3\pi}{4}$ or $\frac{7\pi}{4}$ (because using the unit circle, using those two should cause the sin and cos to cancel out), so that $x = e^{{3\pi}/{4}}$ or $x = e^{{7\pi}/{4}}$, but both of those are clearly far more higher then the x-coordinates of the min and max points in the graph.

How can I find the min and max x-coordinates of this equation? This is a homework problem I've been puzzling over for some time, so either hints or full answers would be appreciated.

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up vote 1 down vote accepted

There are multiple values for $\arctan(-1)$. Note that the full inverse would be given by $-\frac{\pi}{4} + k \pi$ for $k\in \mathbb{Z}$.

In particular, it seems that $\exp(-\frac{\pi}{4})\approx 0.4559$ and $\exp(-\frac{5\pi}{4}) \approx 0.0197$ seems to be the values you're after (by taking $k=0$ and $k=-1$ whereas you took $k=1$ and $k=2$).

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Thanks -- it seems really obvious now. – Michael0x2a Oct 14 '12 at 21:16

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