How can the point of inflection change before the vertical asymptote? I have to draw a graph of a function which seems to have an inflection point AFTER the vertical asymptote.
i.e. f(x) = $\tan^{-1}\left({\frac{x-1}{x+1}}\right)$
Using the quotient rule, I get...
$$f'(x) = \frac{1}{1+\left(\frac{x-1}{x+1}\right)^2}.\frac{(x+1)-(x-1)}{(x+1)^2} $$
Simplifying slightly, I reached...
$$f'(x) = \frac{2}{(x+1)^2+\frac{(x+1)^2(x-1)^2}{(x+1)^2}}$$
Would I be right in thinking this can be simplified further to...
$$f'(x) = \frac{1}{x^2+1}?$$
As technically they are different functions since the first is not defined for "$x= -1$", but the second is.
The problem I came across was when finding the point of inflection. I got the second derivative to be...
$$f''(x) =\frac{-2x}{(x^2+1)^2}$$
When making this equal zero to find the points of inflection, I found it to be -2x = 0, hence x = 0. But the issue is the asymptote is at x = -1. The curve is concave up right up before the asymptote, but apparently is still concave up after the asymptote between x = -1 and x = 0.
Even checking this on Google's graph widget seems to show an inflection point at x = 0 then an awkward line as it approaches the asymptote from the right.
Any ideas on the reason for this or if I've missed something?

Note that your function is equivalent to arctanx - (pi/4)
