Arctangent of a rational function.

$f(x)= \arctan \frac {x}{x-1}$ .

This function is defined on $\mathbb{R}-\{ 1 \}$ .

The function $f$ is the composition of $\arctan$ on a rational function; hence it is differentiable on its domain. We have:

$$\forall x \in \mathbb{R}-{1}, f'(x)= \frac {1}{1+ \left( \frac {x}{x-1} \right) ^2} \cdot \frac {-1}{(x+1)^2}= \frac {-1}{2x^2-2x+1}$$

The denominator never vanishes and is positive at every point of the domain, thus $f$ increases over $(-\infty, 1)$ and over $(1,+\infty)$ .

The function $f'$ is a rational function; hence it is differentiable on its domain. We have:

$$\forall x \in \mathbb{R}-{1}, f''(x)=\frac {2(2x-1)}{(2x^2-2x+1)^2}$$

The second derivative vanishes at $\frac 12$ ; its is negative on the left and positive on the right of this point. Thus the function $f$ is concave on $(- \infty , \frac 12 )$ and on $( \frac 12, 1)$ and $f$ is convex on $(1,+\infty)$ .

Let us now compute the limits of $f$ at the open endpoints of its domain.

Figure: The graph of $f(x)= \arctan \frac {x}{x-1}$ .