Minimization of a function including quadratic and exponential terms Apologies for the simple question. Is it possible to analytically minimize a function such as
$$f(x) = a^x + c x (x-1) \qquad x > 0, \quad 0 < a,c < 1$$
Nothing else can be said about the constants $a$ or $c$. 
 A: Alright, here's my expansion of my comment.
Let's rearrange the equation in VFG's answer like so:
$$\frac{\ln\;a}{2c}=\left(\frac12-x\right)a^{-x}$$
and multiply both sides by $\sqrt{a}$:
$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)a^{\frac12-x}$$
We're close to a form where we can now use the Lambert function, but let's change bases first:
$$\frac{\sqrt{a}\ln\;a}{2c}=\left(\frac12-x\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$
We're much closer now, so we multiply both sides by a factor of $\ln\;a$
$$\frac{\sqrt{a}(\ln\;a)^2}{2c}=\left(\frac{\ln\;a}{2}-x\ln\;a\right)\exp\left(\frac{\ln\;a}{2}-x\ln\;a\right)$$
and then perform the inversion:
$$W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)=\frac{\ln\;a}{2}-x\ln\;a$$
which is now easily solved for $x$:
$$x=\frac12-\frac1{\ln\;a}W\left(\frac{\sqrt{a}(\ln\;a)^2}{2c}\right)$$
Now, the Lambert function has two real branches $W(z)$ (principal branch) and $W_{-1}(z)$; one is real for $z\geq-e^{-1}$, and the other is real only in the interval $[-e^{-1},0)$. If both $a$ and $c$ are in the interval $(0,1)$, the argument of the Lambert function would be much greater than 1, and thus we can dispose of the $W_{-1}$ branch, as well as consider $W(z)$ only for $z\geq 0$, which is true for $a$ and $c$ in the interval $(0,1)$. Thus, the Lambert solution using the principal branch is valid.
A: I think there is no analytical solution to $f^{\prime }(x)=0$
$f^{\prime }(x)=2cx-c+\left( \ln a\right) a^{x}$.
Update: I mean no solution in elementary functions.
