Intuitive reason for why $ \frac{1}{\ln(x)}$ has a change in concavity around $y = -\frac{1}{2}$

Question

The graph for $ y = \frac{1}{\ln(x)}$ has a change in concavity right about where $y=-\frac{1}{2}$ Is there any intuitive reason for this, or way you could know with high school maths (other than calculus)?

Answer 1

Let's consider the behavior of $y$ for all positive real numbers $x$. For $0 < x < 1$, ln $x$ is negative and increasing from $-\infty$ to 0, so $y$ is negative and decreasing from 0 to $-\infty$. You have a vertical asymptote at $x = 1$. For $x > 1$, ln $x$ increases from 0 to $\infty$, so $y$ decreases from $\infty$ to 0. Knowing this and the shape of the graph of ln $x$, it should be easy to graph 1/ln $x$. I'm not exactly sure what you mean by change in concavity though.

Answer 2

Your function has two vertical asymptotes at $x=0$ and $x=1$. The steep is very big and negative near those points.

However, as you go further from $0$ and $1$ and into the middle of the interval, the steepness is small. This means that at some point, the steepness of your graph has to stop getting smaller and start getting larger again. At that point, you have your inflection point.

I am not sure, though, about how much of this is just trying to put calculus ideas in English-like words, but I think you can get the idea that concavity/convexity is a concept that deals with changes in the slope (i.e: is the slope increasing or decreasing?), but first and second derivatives really help to manage this

Answer 3

Before $e^{-2}$ the concavity was upwards (convex)

After $e^{-2}$ the concavity was downwards (concave)

Is this considered a way of noticing it by high school math only from the drawing? Or not?

Your question isn't really clear

Answer 4

Not very intuitive, but a nice alternative view based directly on the definition of $\log$ as an integral: Since the function $x \mapsto x^{-1}$ is convex for positive $x$ it follows that for any positive $a\leq x$ $$\log(x)-\log(a) = \int_a^x \frac{\mathrm{d}t}t \geq (x-a)\cdot \frac{1}{\frac{x+a}2}=2\frac{x-a}{x+a}.$$ For $0<x\leq a$ the reversed inequality holds (with $\leq$ instead of $\geq$). In fact, the function $$\log(a) + 2\frac{x-a}{x+a}$$ is tangent to $\log(x)$ at $x=a$, less than $\log(x)$ for $x>a$, and greater than $\log(x)$ for $0<x<a$. It is a hyperbola with horizontal asymptote $\log(a) + 2$. In particular for $a = e^{-2}$, it “simplifies” to $$\frac{-4}{e^2 x + 1}.$$ Taking reciprocals it follows that the line $-\tfrac14(e^2x+1)$ is tangent to $\log(x)^{-1}$ at $x=e^{-2}$, greater for $x>e^{-2}$, and less for $x<e^{-2}$. In particular, $(e^{-2}, -\tfrac12)$ is an inflection point of $\log(x)^{-1}$.