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:

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

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:

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

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:

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

Now we have:

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

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 $f$ :

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

It follows that

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

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