# 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?

## 2 Answers

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 $f$.

• I'm still confused... Using other graphing sites, I see it goes up to about y = π/2 then jumps down to y = −π/2 at x = -1. Is this a jump discontinuity? Since the function is not defined for x = -1? The graphs always seem to draw lines between y = π/2 and -π/2, but surely that is incorrect also... since there are no real values for which f'(x) = 0. – Wolff Feb 10 '14 at 18:04
• Yes, the function is undefined at $x=-1$ but has left-hand limit $\pi/2$ and right-hand limit $-\pi/2$. So there is a jump, but strictly speaking there is no discontinuity, since that word should only be applied to points in the function's domain of definition where the function is not continuous. (This function is actually continuous, since it is continuous at every point where it is defined. But of course, because of the jump, there is no way to assign a value $f(-1)$ which would make $f$ continuous on the whole real line.) – Hans Lundmark Feb 11 '14 at 9:50

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

• Not for $x<-1$. – Hans Lundmark Feb 10 '14 at 6:46
• @David: Perhaps you should spell out how this attempts to answer the Question. Good one-line Answers, particularly to lengthy Questions, are quite rare on Math.SE. – hardmath Feb 10 '14 at 12:23