Suppose if I'm required to draw the graph of $$y=\frac{1}{\ln\lvert x\rvert}$$ through transformations.

I first draw the graph of $y=\ln(x)$.Then replacing $x$ by $\lvert x\rvert$ ,through transformations I trace the mirror image of the graph of $y=\ln(x)$ on the other side of $y$ axis to get the graph of $y=\ln\lvert x\rvert$.

Now since $\frac{1}{\ln\lvert x\rvert}$ is the inverse of $\ln\lvert x\rvert$, I finally draw the required graph by taking the reflection of the graph of $y=\ln\lvert x\rvert$ about the line $y=x$.

This should've given me the correct graph , but upon seeing the actual graph using a graph calculater , my graph turns out to be completely different.

Can someone just tell if there is any flaw in my logic or if I'm doing anything wrong ? What's the possible way to trace this curve ?

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    $\begingroup$ Note that the inverse function is under composition, not multiplication. The inverse function of $ln(x)$ is $e^x$, not it's reciprocal. $\endgroup$ – CyclotomicField Nov 23 '17 at 13:42
  • $\begingroup$ Yes @Arthur's answer cleared that doubt for me. Can you explain to me graphically what he means in the last paragraph of his answer ? $\endgroup$ – Tanuj Nov 23 '17 at 13:47
  • $\begingroup$ He's describing the poorly behaved nature of the transformation $f(x) \rightarrow \frac{1}{f(x)}$ which isn't really a "reflection" in the colloquial sense. The only situation where I know that $\frac{1}{x}$ is well behaved on the graphs of functions is on the Riemann Sphere where it inverts the sphere along its equator. $\endgroup$ – CyclotomicField Nov 23 '17 at 14:13

There are two different kinds of inverses when talking about functions, because there are two very different operations that both give rise to inverses. This is confusing, and there is no way around it except to always be conscious, and perhaps even specific, about which inverse you are talking about.

One operation, which I suspect is the one you're thinking of, is composing functions. In that case, the inverse of a function $f(x)$ is a function $g(x)$ such that $f(g(x)) = g(f(x)) = x$ (in the context of composing functions, $x\mapsto x$ is the neutral function). We usually write $f^{-1}(x)$ for this inverse. The graph transformation is simply reflection about the line $x = y$.

The other operation, which is the one that's relevant here, is pointwise multiplication. In this case, the inverse function of $f(x)$ is a function $g(x)$ such that $f(x)\cdot g(x) = 1$ (in the context of multiplication, $x\mapsto 1$ is the neutral function). We usually write $\frac1{f(x)}$ for this inverse. It is sometimes called the "reciprocal", however, and this would clear some confusion.

I don't know exactly what kind of transformations you're talking about. But if I had to describe the visual transformation from the graph of $f(x)$ to the graph of $\frac1{f(x)}$, I would say that it is a form of reflection. Everything above the $x$-axis is "reflected" about the line $y = 1$, and everything below the $x$-axis is "reflected" about the line $y = -1$ in such a way that

  • Which side of the $x$-axis a point is on is not changed
  • Points very close to the $x$-axis end up very far away
  • Points very far away from the $x$-axisend up very close

Simple example, using $f(x) = x$:

Graph transformation

The blue curve is a distorted reflection of the green line about the red line. Points close to the $x$-axis end up far away, and points far away from the $x$-axis end up close to the $x$-axis.

More complicated example, with $f(x) = 2\sin(x)$:

enter image description here

Here we again see this distorted reflection about the red and orange lines at work. Also, note that whenever the original function is above the $x$-axis, the reflection is above the $x$-axis, and whenever the original function is below the $x$-axis.

  • $\begingroup$ I understand somewhat what you mean .Can you illustrate the point you made in the last paragraph of your answer using graphs ? Its a bit confusing what you really mean. $\endgroup$ – Tanuj Nov 23 '17 at 13:46
  • $\begingroup$ @Tanuj I added two examples you can take a look at. $\endgroup$ – Arthur Nov 23 '17 at 14:00
  • $\begingroup$ Sorry @Arthur , could you explain what you mean by that ? "Points close to the x x -axis end up far away, and points far away from the x x -axis end up close to the x x -axis" ? $\endgroup$ – Tanuj Nov 23 '17 at 14:03
  • $\begingroup$ @Tanuj I'm just trying to explain how the reflection distorts things. If, at some point, the graph of $f(x)$ is far away from the $x$-axis, then the graph of $\frac 1{f(x)}$ will be close to the $x$-axis, and vice versa. You can see this, for instance, the top of the green graph in the second example. That's the furthest the green graph gets from the $x$-axis. At that point, the blue graph is as close to the $x$-axis as it gets. As the green graph comes back down towards the $x$-axis, the blue graph moves up and away from the $x$-axis. $\endgroup$ – Arthur Nov 23 '17 at 14:06
  • $\begingroup$ There really isn't much more I can do to help you through text and images. You just have to look at them until you see how the reflection works, or try a few examples yourself. $\endgroup$ – Arthur Nov 23 '17 at 14:07

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