Minimize exponential function over an interval I want to minimize the function
$$f(x) = e^{ax^2} + e^{b(1-x)}$$
with respect to $x$ (where $a$ and $b$ are constants), subject to the constraint that $0 \leq x \leq 1$.
I know that $x = 0$ and $x = 1$ are both critical points, but I am interested in finding the potential minimum which lies between the boundary values.
Surely the first step is to set the derivative equal to $0$, yielding
$$0 = f'(x) = 2axe^{ax^2} - be^{b(1-x)}$$
The minimizing $x$ must solve $2axe^{ax^2} = be^{b(1-x)}$, but I'm not sure how to "use" the constraint $0 \leq x \leq 1$ to do so...
Note:  In my specific problem, I have $a = -\frac{N}{16}$ and $b = -\frac{N}{4} (\frac{1}{2} - p)^2$, where $N$ is a constant positive integer, and $p$ is a constant with $0 \leq p < \frac{1}{2}$.  Ideally, we will be able to express a minimizing $x$ in terms of $N$ and $p$.
 A: If $b \neq 0$ and $2ax \geq b$ for all $0 \leq x \leq 1$ then $ax^2-bx+b$ is increasing on $0 \leq x \leq 1$ and $\frac{2a}{b} e^a \leq 1$.  These statements imply that
$$
\frac{2ax}{b}e^{ax^2 - bx + b} < 1
$$
for $0 \leq x < 1$, which means that $f'(x) \neq 0$ there.  Thus you may conclude that your minimum occurs at one of the endpoints.
What restrictions are there on $a$ and $b$?  Perhaps similar arguments exist for other specific cases.
Edit: As I will probably not have enough time to work on this in the next 7 days I will sketch here what I was thinking about in the hopes it will benefit others.  My goal was to put a lower bound on the minimum of the function $f(x) = e^{ax^2} + e^{b(1-x)}$ using the constraints for $a$ and $b$ in the note at the end of the question.
For $y > 0$, replacing
$$
e^{ax^2} \hspace{0.5cm} \text{with} \hspace{0.5cm} \frac{1}{y+e^{-ax^2}}
$$
or
$$
e^{b(1-x)} \hspace{0.5cm} \text{with} \hspace{0.5cm} \frac{1}{y+e^{-b(1-x)}}
$$
in $f(x)$ will yield a function $f_y(x)$ which is less than $f(x)$.  Depending on which choice was made, $f_y'(x)$ will be either nonnegative or nonpositive for $y$ large enough.
To determine how large it needs to be, we start by solving $f_y'(x) = 0$ for $y$, which I believe will always reduce to a quadratic equation.  These two roots of $f_y'(x) = 0$ are functions of $x$ which I think can always be made less than a sum of two exponentials.  This sum will be bounded (since x is bounded), so we would like to find an upper bound (to elminate the $x$ dependence).  Here I would have tried to use the inequality $e^x \leq 1/(1-x)$, which holds for $x<1$, to bound this sum of exponentials above by a sum of rational functions.  Thus we end up wanting to solve something like
$$
y > \frac{1}{1-p(x)} + \frac{1}{1-q(x)}
$$
where $p(x)$ and $q(x)$ are quartic (or less) and $0 \leq x \leq 1$.  An upper bound of the right-hand side is theoretically possible to write down.  Taking $y$ larger than this will make $f_y(x)$ monotonic, and hance the minimum of $f(x)$ will be greater than the lowest endpoint of $f_y(x)$, whose height can (theoretically) be calculated.
A: Alway plug in the boundary values. Optima occur among critical points; the oundary values are, by definition critical. Here $f(0)= e^b$, while $f(1)=e^a.$ To find your other critical points, try dividing one side by the other.
