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So the whole problem that I'm trying to solve simply states: analyse the following function and sketch its graph:

$$f(x) = -x + \ln(\frac{|x|-1}{x})$$

The thing that's problematic here is determining the sign of the function and its zeroes. Solving the equation:

$$-x + \ln(\frac{|x|-1}{x}) = 0$$ seems rather difficult, if not impossible. So I came up with an idea using which I got an approximately correct answer but I'm not sure how "legal" it is.

So the domain of the functions is:

$$ Df: (-1, 0) \cup (1, + \infty)$$

Let's first look at the left side, the one that goes from -1 to 0. $$\lim_{x \to -1^+}f(x) = \lim_{x \to -1^+}(-x + \ln(\frac{-x-1}{x}))= -1+\lim_{x \to -1^+}{\ln(\frac{-x-1}{x})} = -1 + \ln(\frac{1^- - 1}{-1^+}) = -1 + \ln(\frac{0^-}{-1^+}) = -1 + \ln(0) = -1 - \infty = - \infty$$

I did the similar thing for the right bound of the interval

$$\lim_{x \to 0^-}{f(x)} = \lim_{x \to 0^-}{\ln(\frac{-x-1}{x})} = \ln(\frac{0^+-1}{0^-}) = \ln(\frac{-1}{0^-}) = \ln(\infty) = \infty$$

So basically I got: $$ \text{as $x \to -1^+$, $f(x) \to -\infty$<0} \\ \text{as $x \to 0^-$, $f(x) \to +\infty$>0} $$

And since the function is obviously continuous on the interval $$(-1, 0)$$ could I somehow use the Bolzano-Cauchy mid-value theorem to state that $$\forall C \in (-\infty, +\infty) \quad \exists c : f(c) = C $$ And then simply set $$ C = 0$$ thereby concluding that the function must intersect the x axis on $$ (-1, 0) $$ at the same time finding that before that value it's negative and after that value it's positive.

Now this is actually correct, I mean I've looked at the graph of this function and it does indeed intersect the x axis in between -1 and 0. But I'm wondering how legal was my application of the Bolzano-Cauchy theorem here, since the theorem clearly requires that the function be continuous on a closed interval and it also requires that the value at one side of the interval be less than the value at the other side of the interval; but in my case the function isn't even defined at those points.

So did I just get lucky here or is there some justification (some other theorem maybe) for the way I've gotten to my solution.

Also if anyone has any other ideas how to figure out the sign and zeros of this function, that would also be nice.

Thanks.

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  • $\begingroup$ Yeah, there's a minor mistake that I've made in the first limit but it does not affect the result $\endgroup$ – Koy Sep 17 '18 at 12:22
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Your reasoning is correct. Notice that, by definition, $\lim_{x \to 0^-} f(x) = +\infty$ means that for all $M$, there exists a $\delta > 0$ such that $0 < 0 - x < \delta$ implies that $f(x) > M$. On the other hand, $\lim_{x \to -1^+} f(x)$ means that for all $N$, there exists an $\eta > 0$ such that for $0 < x + 1 < \eta$ implies that $f(x) < - N$.

Now, fix $C$ and choose $N = M = 2|C| + 1$. Then, choosing the $\delta$ and $\eta$ from the definitions gives us that $f(-1 + \eta/2) < -2|C|$ and $f(- \delta/2) > 2|C|$. By applying the Bolzano-Cauchy theorem (better known as the Intermediate Value Theorem) to the interval $[-1 + \eta/2, -\delta/2]$, we see that there must exist a $c$ with $-1 < -1 + \eta/2 < c < - \delta < 0$ such that $f(c) = C$.


Do notice, however, that you need a little more information to conclude that $f$ is positive to the right of $c$ and left to the left of $c$: this means that $f$ is increasing over $(-1, 0)$ (and strictly increasing near $0$). You can justify this by taking there derivative of $f$ and noticing that it is positive over this interval.

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  • $\begingroup$ Thanks for the reply. So basically, intuitively speaking, since f is not defined at -1 and 0 I can, kind of, "shift" the left end of the interval by a very small value to the right and also shift the right end of the interval by a very small value to the left; and in those new edges the function will be defined, giving me the liberty to use the intermediate-value theorem. I've thought of something like this but couldn't quite formally write it. Thanks again. $\endgroup$ – Koy Sep 17 '18 at 16:50
  • $\begingroup$ @Koy, yes, that is the idea. The limits at the endpoints guarantee that this shifting procedure will work 'as if' you were applying it at the endpoints where the function value diverges to $\pm \infty$. $\endgroup$ – Strants Sep 17 '18 at 16:54

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