Finding Concavity in Curves Suppose
$$\frac{d^2y}{dx^2} = \frac{e^{-t}}{1-e^t}.$$
After finding the second derivative, how do I find concavity? It's much easier to solve for $t$ in a problem like $t(2-t) = 0$, but in this case, solving for $t$ seems more difficult. Does it have something to do with $e$? $e$ never becomes $0$, but at what point is the curve upward or downward?

From the t on the RHS I assume that this comes from a pair of parametric equations where you were given x(t) and y(t) and that you got the second derivative by calculating dm/dx=m'(t)/x'(t) where m(t)=dy/dx=y'(t)/x'(t). 
In any case, you are right that the curve will be concave up at points where the second derivative is positive and concave down where it is negative. 
You are also right that the exponentials are never zero. e itself is just a specific positive constant and so raising it to any (positive or negative) power always gives a positive result. So the numerator in your expression is never zero and the only way the whole thing can change sign is when the denominator does. This happens when e^t=1, which means that t must be zero. Since e^t is an increasing function of t it must be bigger than 1 for t>0, and so at those points the second derivative is negative and the curve is concave down. But for the points with negative t, e^t is less than 1 (eg e^(-2)=1/e^2), so at these points the second derivative is positive and the curve is concave up.
