Rational function of an exponential.

$ f(x) =\frac {e^x -1}{e^x-2} $ .

This function is defined on $ \mathbb{R}-\{ \ln 2 \}$ .

The function $ f$ is the composition of a rational function on the exponential; thus, it is differentiable on its domain. We have:

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

This derivative is never equal to 0, and for any $ x \in \mathbb{R}-\{ \ln 2 \},
; f'(x)<0$ . Hence, the function $ f$ decreases on $ (- \infty, \ln 2 )$ and on $ (\ln 2 , + \infty )$ .

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

$\displaystyle \underset{x \rightarrow - \infty}{\lim} e^x =0 \Longrightarrow \b...
...d{cases} \Longrightarrow \underset{x \rightarrow - \infty}{\lim} f(x)=\frac 12.$    

Therefore, the line whose equation is $ y=1/2$ is an asymptote to the graph of $ f$ .

As the exponential function with basis $ e$ has $ +\infty$ as its limit at $ +\infty$ , we need some lagberaic work. We have:

$\displaystyle f(x)=\frac {e^x -1}{e^x-2} = \frac {e^x(1-e^{-x})}{e^x(1-2e^{-x})} =\frac {1-e^{-x}}{1-2e^{-x}}.$    

Now we have:

$\displaystyle \underset{x \rightarrow +\infty}{\lim} e^{-x} =0 \Longrightarrow ...
...) =1 \end{cases} \Longrightarrow \underset{x \rightarrow +\infty}{\lim} f(x)=1.$    

Therefore, the line whose equation is $ y=1$ is an asymptote to the graph of $ f$ .

At $ \ln 2$ , we need to compute one-sided limits:

\begin{align*}\begin{cases}\underset{x\underset{<}{\rightarrow} \ln 2}{\lim} (e^...
...row \underset{x\underset{<}{\rightarrow} \ln 2}{\lim} f(x)=- \infty.\end{align*}    
\begin{align*}\begin{cases}\underset{x\underset{>}{\rightarrow} \ln 2}{\lim} (e^...
...row \underset{x\underset{<}{\rightarrow} \ln 2}{\lim} f(x)=+ \infty.\end{align*}    

The line whose equation is $ x=\ln 2$ is an asymptote to the graph of $ f$ .

Let us now compute the second derivative of $ f$ :

$\displaystyle \forall x \in \mathbb{R}-\{ \ln 2 \}, \; f''(x)= \frac {e^x \; ( e^x +2)}{(e^x-2)^3}.$    

It follows that

\begin{displaymath}\begin{cases}x< \ln 2 \Longrightarrow f''(x)<0  x> \ln 2 \Longrightarrow f''(x)>0 \end{cases}\end{displaymath}    

Therefore, $ f$ is concave over $ (- \infty, \ln 2 )$ and convex over $ (\ln 2 , + \infty )$ ; it has no point of inflexion.

We display the graph of $ f$ in Figure 4.

Figure 4: The graph of $ f(x) =\frac {e^x -1}{e^x-2} $ .
\begin{figure}\mbox{\epsfig{file=graph4.eps,height=6cm}}\end{figure}

Noah Dana-Picard 2007-12-28