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:

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

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:

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

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}$ .
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Noah Dana-Picard 2007-12-28