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?
Google is plotting the graph incorrectly. There is no asymptote. (The function cannot tend to infinity, since it is bounded between $-\pi/2$ and $\pi/2$.)
In fact, since $f'(x)=1/(1+x^2)$ for $x<-1$ and for $x>-1$, it follows that $f(x)=\arctan x + C_1$ for $x<-1$ and $f(x)=\arctan x+C_2$ for $x>-1$, but $C_1$ may not be the same as $C_2$. If you know what the graph $y=\arctan x$ looks like, you can just plug in some values to determine $C_1$ and $C_2$, and then use that to draw the graph of $f$.