Intuitive reason for why $ \frac{1}{\ln(x)}$ has a change in concavity around $y = -\frac{1}{2}$ 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)?
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 $0&lt;x\leq a$ 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 $x&gt;a$, and greater than $\log(x)$ for $0&lt;x&lt;a$. 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 $x&gt;e^{-2}$, and less for $x&lt;e^{-2}$. In particular, $(e^{-2}, -\tfrac12)$ is an inflection point of $\log(x)^{-1}$.
Let's consider the behavior of $y$ for all positive real numbers $x$. For $0 &lt; x &lt; 1$, 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 $x &gt; 1$, 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.