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Problem:

Find a curve whose slope at each point $(x,y)$ equals the reciprocal of the $x$-value if the curve contains the point $(e,-3)$.

If we call the equation of the curve $f(x)$, then $f'(x)=\frac{1}{x}$. $\int \frac{1}{x} \, dx = \ln|x|+C$, so $f(x)= \ln|x|+C$. $f(e)=-3$ so $f(x)= \ln|x|-4$. What I am confused by is that my book's answer is $f(x)= \ln x-4$ (there are no absolute value bars around $x$). If there are no bars, that means the bars are uneccessary so $x$ must be greater than zero. But what part of the problem indicates that $x$ must be greater than zero?

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Since $\frac{1}{x}$ is discontinuous (update: Hans Lundmark's comment suggests using "undefined" instead) at $x = 0$, there are actually $2$ curves involved, one where $x \lt 0$ and another where $x \gt 0$. Since the point in question of $(e,-3)$ has $x = e \gt 0$, it's only this positive $x$-values curve which is being asked about. Thus, there's no need for absolute value signs in this case. Nonetheless, I believe the book should still have used absolute values, even if they're not strictly necessary.

Update: After reading the comments below and on further reflection, I now believe it's best to not use absolute values, as this helps to implicitly show the solution is only for $x \gt 0$. I suspect this is a major reason why the book did not use absolute values in their answer.

John Omielan
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  • How are there 2 curves? The integral of $\frac{1}{x}$ can still be continuous at $x=0$. –  Jul 30 '19 at 01:24
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    No, it is not continuous. $\ln(x) \to -\infty$ as $x \to 0+$. If you want a connected curve, the answer is $\ln(x)-4$ for $x > 0$. There is no reason to include a curve $\ln(-x)+c$ for $x<0$, which is not connected to the first curve; and if you wanted to include one, there's no reason to have $c=-4$. – Robert Israel Jul 30 '19 at 01:26
  • @user532874 Note the question says the curve's slope is the reciprocal of the $x$-value. Thus, as $x \to 0^{+}$, the slope will approach positive infinity, with $\ln x$ going to $-\infty$, as Robert Israel stated. Thus, the curve is discontinuous at $x = 0$, so for $x \lt 0$, if you wish to consider that region, it's a different "curve" (i.e., not connected to the first one, as Robert's comment indicates), even if you have one defined function which encompasses both curves. – John Omielan Jul 30 '19 at 01:30
  • @RobertIsrael Thanks for your comment & explanation. You did a better job of it than I could. – John Omielan Jul 30 '19 at 01:31
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    This is one illustration of why I consider it unfortunate that calculus students are taught $\int \frac{1}{x} dx = \ln |x| + c$. It really should be $\ln(x) + c_1$ for $x > 0$ and $\ln(-x) + c_2$ for $x < 0$, where there is no connection between $c_1$ and $c_2$. – Robert Israel Jul 30 '19 at 01:34
  • I do see now that $\ln x$ is not continuous at zero but I still don’t see why there would be 2 curves. At $x<0$ there are no defined points for $\ln x$. –  Jul 30 '19 at 01:50
  • @user532874 The curve for $x \gt 0$, & for $x \lt 0$ if you choose to define for that domain, are not connected. Thus, depending on how you define "curve", they are $2$ separate curves. If all you're concerned about is $x \gt 0$, then you can use $\ln x$. However, if you wish to also consider $x \lt 0$, then you need to use something like $\ln |x|$, or $\ln(-x)$ for $x \lt 0$. As Robert's comment states, although it's taught $\int \frac{1}{x}dx = \ln |x| + c$, there are really $2$ curves which should be treated separately, e.g., the $c$ values don't need to match between the $2$ curves. – John Omielan Jul 30 '19 at 01:56
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    @user532874 In particular, the condition $f(e)=-3$ pins down the location of the curve for $x>0,$ but not the curve for $x<0.$ The curve could pass through $(-1,0)$ or $(-1,-17)$ or $(-1,9987)$ or infinitely many points all of which force the $x<0$ part to be different from all the other choices. Which one do you choose? The problem statement gives no advice on that, so it seems reasonable that the solution does not either. – David K Jul 30 '19 at 02:13
  • Nitpick: It's best not to use the word discontinuous when referring to points where the function is undefined, since it's strange to say that continuous(!) functions like $1/x$ or $\ln x$ have discontinuites. See here: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark Jul 30 '19 at 11:12