Square root and rational function.

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

For this function to be defined, we need that $ x \neq -1$ and $ \frac {x-1}{x+1} \geq 0$ .Thus, this function is defined on $ (-\infty, -1) \cup [1,+\infty )$ .

The function $ x \mapsto \frac {x-1}{x+1}$ is a rational function, thus it is differentiable on its domain; the composition of this function with the square root function is differentiable where the rational function does not vanish. Finally $ f$ is the product of this composition with a linear function, hence $ f$ is differentiable on $ (-\infty, -1) \cup (1,+\infty )$ . Let us compute the first derivative:

$\displaystyle \forall x \in (-\infty, -1) \cup (1,+\infty ), \; f'(x)$ $\displaystyle = \sqrt{ \frac {x-1}{x+1}} + x \cdot \frac {1}{2\sqrt{ \frac {x-1}{x+1}}} \cdot \frac {2}{(x+1)^2}$    
$\displaystyle \quad$ $\displaystyle = \frac {x^2 + x - 1}{\sqrt { \frac {x - 1}{x + 1}} \cdot (x + 1)^2}$    

This first derivative vanishes if, and only if:

$\displaystyle x^2+x-1=0$    


$\displaystyle x=\frac {-1-\sqrt{5}}{2}$    or $\displaystyle x=\frac {-1+\sqrt{5}}{2}$    

Only the first one belongs to the domain of $ f$ . Now for $ x$ in the domain of differentiability, we have:

$\displaystyle f'(x)>0 \Longleftrightarrow x < \frac {-1-\sqrt{5}}{2}$    or $\displaystyle x>1$    

The function $ f$ increases on $ (-\infty, (-1-\sqrt{5})/2 )$ and on $ (1,+\infty)$ , and it decreases on $ ( \; (-1-\sqrt{5})/2, -1 \; )$ .

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

\begin{align*}\begin{cases}\underset{x \rightarrow -\infty}{\lim} \frac {x-1}{x+...
... 1  \underset{x \rightarrow -\infty}{\lim} x = -\infty \end{cases}\end{align*} $\displaystyle \Longrightarrow \underset{x \rightarrow -\infty}{\lim} f(x)=-\infty$    
\begin{align*}\begin{cases}\underset{x \rightarrow +\infty}{\lim} \frac {x-1}{x+...
... 1  \underset{x \rightarrow +\infty}{\lim} x = -\infty \end{cases}\end{align*} $\displaystyle \Longrightarrow \underset{x \rightarrow +\infty}{\lim} f(x)=-\infty$    
\begin{align*}\begin{cases}\underset{x \underset{<}{\rightarrow} -1}{\lim} (x-1)...
...et{x \underset{<}{\rightarrow} -1}{\lim} (x+1)=0^{-1}  \end{cases}\end{align*} $\displaystyle \Longrightarrow \underset{x \underset{<}{\rightarrow} -1}{\lim} \...
...y \Longrightarrow \underset{x \underset{<}{\rightarrow} -1}{\lim} f(x)=-\infty.$    

Therefore the line whose equation is $ x=-1$ is an asymptote to the graph $ \mathcal{C}$ of $ f$ .

We shall now look for oblique asymptotes:

$\displaystyle \underset{x \rightarrow +\infty}{\lim} \frac {f(x)}{x} = \underset{x \rightarrow +\infty}{\lim} \sqrt{\frac{x-1}{x+1}} = 1$    

Now we can show that

$\displaystyle \underset{x \rightarrow +\infty}{\lim} (f(x)-x)= -1$    

This means that the line whose equation is $ y=x-1$ is an oblique asymptote to the graph $ \mathcal{C}$ of $ f$ . Actually, the same line is an asymptote to $ \mathcal{C}$ for $ x$ in a neighborhood of $ -\infty $ (please check this!).

We can now draw the variation table of $ f$ (where $ a=(-1-\sqrt{5})/2$ and $ b=f(a)$ ):


Let us look for a point of inflexion. By arguments similar to those we used for differentiating $ f$ , we prove that $ f'$ is differentiable on $ (-\infty, -1) \cup (1,+\infty )$ and we have:

$\displaystyle \forall x \in (-\infty, -1) \cup (1,+\infty), \; f''(x)=\frac {(x-2)\sqrt{\frac {x-1}{x+1} }}{(x-1)^2 \; (x+1)^2}.$    

The second derivative vanishes at 2, it is negative for $ x<2$ and positive for $ x>2$ ; thus, $ f$ is concave on $ (- \infty , -1)$ and on $ (1,2)$ and it is convex on $ (2,+\infty)$ . At 2 the function $ f$ has a point of inflexion.

Figure 2: The graph of $ f(x) = x \sqrt { \frac {x-1}{x+1}}$ .

Noah Dana-Picard 2007-12-28