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I'm having problems in understanding few things about this function, also because some of my calculations do not match the plot. $$f(x) = \dfrac{-2}{5x-\ln\vert x \vert}$$

Here is what I did.

First of all, the domain is $\Omega: x\in (-\infty, x^*)\cup(x^*, 0)\cup(0, +\infty)$ where $x^*$ is the point that makes $5x - \ln\vert x \vert = 0$. I understood through a very easy analysis that it's negative.

The domain is a union of open sets to there is no garancy we have max or min but we go on the same. I calculated these behaviours:

$$\lim_{x\to \pm\infty} f(x) = 0^{\mp}$$

$$\lim_{x\to 0^{\pm}} f(x) = 0$$

$$\lim_{x\to x^{*+}} f(x) = -\infty$$ $$\lim_{x\to x^{*-}} f(x) = +\infty$$

So thanks to this I sketched the plot a bit, understanding that there is a vertical asymptote at $x^*$ and a horizontal asymptote at zero. At $x = 0$ the function is continuos (removable singularity) but not differentiable (it might be a cusp point), for

$$f'(x) = \dfrac{2}{(5x-\ln\vert x\vert)^2}\left(5 - \dfrac{1}{\vert x \vert}\right)$$

Where we see the non differentiability at $x^*$ and at $0$. Also $f(x)$ is not continuous at $x^*$.

Troubles start now, where I have to study the monotonicity of the function. Splitting it into positive and negative $x$ I have

$$f'(x) = \begin{cases} \dfrac{2}{(5x-\ln(x))^2}\left(5 - \dfrac{1}{x}\right) \qquad & x > 0 \\ \dfrac{2}{(5x-\ln(-x))^2}\left(5 + \dfrac{1}{x}\right) \qquad & x < 0 \end{cases} $$

When studying $f'(x) > 0$ I recognize the left term is always positive, but the term in the round brackets in the right changes. I get for $x > 0$

$$f'(x) > 0 \rightarrow x > 1/5 $$

And the function is decreasing for $0 < x < 1/5$. This matches the plot.

But for $x < 0$ I get

$$f'(x) > 0 \rightarrow x\in(-\infty, -1/5)$$

And it's decreasing for $-1/5 < x < 0$

The problems here are:

-- The function is actually increasing for $x\in (-\infty, x^*)$ and for $x\in(x^*, 0)$

-- There is no decreasing part for $x < 0$

-- Also the limit for $f(x)$ at zero seems like to not be zero...

Hence here I got stuck. Can someone please help? Thank you!

enter image description here

enter image description here

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    $\begingroup$ What's happening near $x=0$ shows that you should not have such faith in computer graphing programs. Your formula for the derivative is wrong, however. Be careful differentiating $\ln(|x|)$. $\endgroup$
    – Ted Shifrin
    Commented Dec 12, 2022 at 21:57
  • $\begingroup$ @TedShifrin Isn't it $$(\ln\vert x\vert)' = \dfrac{1}{x}\text{sgn(x)} = \dfrac{1}{x}\cdot \dfrac{\vert x \vert}{x} = \dfrac{1}{\vert x\vert}$$? $\endgroup$
    – Heidegger
    Commented Dec 12, 2022 at 21:59
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    $\begingroup$ You mean $\dfrac1{|x|}\text{sgn}(x)$? Check it by cases. Notice that $\ln|x|$ is most definitely decreasing on $x<0$. $\endgroup$
    – Ted Shifrin
    Commented Dec 12, 2022 at 22:02
  • $\begingroup$ @TedShifrin Ahhh, gosh! You're right. $\ln(f)$ "becomes" $f'/f$ hence at the denominator there is $\vert x \vert$. Ha, thank you! $\endgroup$
    – Heidegger
    Commented Dec 12, 2022 at 22:03
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    $\begingroup$ Yes. It is not plotting enough points near enough to $x=0$. If you manually force it to use more points to plot, the cusp moves closer and closer to $y=0$. $\endgroup$
    – Ted Shifrin
    Commented Dec 12, 2022 at 22:15

1 Answer 1

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The problem is when you compute the derivative of $\ln(\vert x \vert)$. Note that the following equality is false

$$ \frac {d}{dx} \ln(\vert x \vert) = \frac{1}{\vert x \vert}$$

Indeed the derivative of an even function cannot be even ! Rather by using the chain rule we can double check that

$$ \frac {d}{dx} \ln(\vert x \vert) = \frac{1}{x}.$$

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  • $\begingroup$ Can you elaborate on the derivative? I have always been told that the derivative of absolute value is signum $$(\ln\vert x\vert)' = \dfrac{1}{x}\text{sgn(x)} = \dfrac{1}{x}\cdot \dfrac{\vert x \vert}{x} = \dfrac{1}{\vert x\vert}$$ $\endgroup$
    – Heidegger
    Commented Dec 12, 2022 at 22:02
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    $\begingroup$ It is. So you're messing up the chain rule. $\endgroup$
    – Ted Shifrin
    Commented Dec 12, 2022 at 22:03
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    $\begingroup$ @Numb3rs chain rule is $(f(g))' = g'\times f'(g)$ where $g'$ is the sign, $f' = 1/x$. So $f'(g) = 1/\vert x \vert$. That is $$ d/dx \ln(\vert x\vert ) = \frac{1}{\vert x \vert } sgn(x)=\frac{1}{\vert x \vert } \times \frac{\vert x \vert }{x} = \frac{1}{x}.$$ $\endgroup$
    – Digitallis
    Commented Dec 12, 2022 at 22:08

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