Is this function unimodal? Consider the real function:
$$f\left(\xi\right)=\frac{1-\xi}{\xi+c\mathrm{e}^{k/\xi}\left(a+\xi-a\xi\right)},\quad0\le\xi\le1$$
where $a,c,k$ are positive parameters. For all combinations of parameters I have tried, this function is always unimodal (I plotted it). I have not been able to prove this because the derivative is quite messy. Is there a simpler way?
On the other hand, by considering expansions near $\xi \approx 0$ and $\xi \approx 1$, I was able to show that this function is increasing near $\xi \approx 0$, and decreasing near $\xi \approx 1$.
 A: Fix $a,c,k > 0$.

Let $f\colon (0,1] \to \mathbb{R}$ be defined by
$$f(x)=\frac{1-x}{x+c\,\mathrm{e}^{k/x}(a+x-ax)}$$
The goal is to show that $f$ is unimodal.

Let $S=\{x \in (0,1)\mid f'(x) = 0\}$.

Note that the denominator of $f$ is positive for all $x$, since 
$$a + x -ax = a(1-x) + x > 0$$

Thus, we have 
$$f(x) > 0\;\;\text{if}\;\,0 < x < 1,\;\,\text{and}\;f(1)=0$$
Next, note that
$$\lim_{x=0^{+}}f(x) = \lim_{x=0^{+}}\frac{1}{ac\,e^{k/x}} = \frac{1}{\infty} = 0$$
It follows that $f$ realizes a global maximum value for some $x \in (0,1)$.

Thus, $f'(x)$ has at least one zero in $(0,1)$, hence $S$ is nonempty.

To show that $f$ is unimodal, it remains to show that $S$ can't have more than one element.

Computing $f'(x)$, setting it to zero, then isolating $e^{k/x}$, we get the equation
$${\large{e^{k/x}}}= \frac{(1/c)x^2}{(ka-k-1)x^2+(k-2ka)x+ka}$$
which must hold at $x=s$, for all $s \in S$.

Thus, for $s \in S$, we must have $h(s) > 0$, where 
$$h(x) = (ka-k-1)x^2+(k-2ka)x+ka$$

Note that $h(0) = ka > 0$, and $h(1) = (ka-k-1)+(k-2ka)+ka=-1 < 0$.

Let $E = \{x \in \mathbb{R} \mid h(x) > 0\}$, and let $D = E \cap (0,1)$.

Of course $D,E$ are open sets.

Since $h$ is continuous, and $h(0) > 0$, it follows that $D,E$ are nonempty.

Claim $D$ is an open interval.$\;$Consider $3$ cases . . .

Case $(1)\,$:$\;ka-k-1 < 0$.

Then $h$ is a quadratic polynomial with negative leading coefficient, hence, since $E$ is nonempty, it follows that $E$ is an open interval, so $D$ is also an open interval.

Case $(2)\,$:$\;ka-k-1 = 0$.

Then $ka = k + 1$, hence $h(x) = (-k-2)x + (k+1)$, hence, since $-k-2 < 0$, $h$ is a polynomial of degree $1$. It follows that $E$ is an open interval, so $D$ is also an open interval.

Case $(3)\,$:$\;ka-k-1 > 0$.

Then $h$ is a quadratic polynomial with positive leading coefficient, hence, since $h(0) > 0$, and $h(1) < 0$, $h$ must have roots $r_1,r_2$, where $0 < r_1 < 1$, and $r_2 > 1$. It follows that $D = (0,r_1)$, so $D$ is an open interval.

Thus, in all $3$ cases, $D$ is an open interval.

Let $g\colon D \to \mathbb{R}$ be defined by
$$g(x)=\frac{x^2}{h(x)}$$
Since $h(x) > 0$ on $D$, it follows that $g$ is differentiable on $D$.

Computing $g'(x)$, we get
$$
g'(x) =
\frac
{(xk)\bigl(2a(1-x)+x\bigr)}
{
c\left(h(x)^2\right)
}
$$
which is positive, since the numerator and denominator are both positive.

Thus, $g$ is strictly increasing on $D$.

But $e^{k/x}$ is strictly decreasing on $D$.

It follows that the equation $e^{k/x} = g(x)$, which holds at $x=s$, for all $s \in S$, has at most one solution in $D$.

But $s \in S$ implies $s \in D$, hence $S$ has at most one element.

Therefore $f$ is unimodal.
