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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?

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You want to figure out where the second derivative is positive and where it is negative. Since the second derivative is continuous wherever it is defined (the exponential is continuous, and so is $1-e^t$, and you have a quotient of continuous functions), the second derivative can change from positive to negative or from negative to positive only at points where it is either equal to $0$, or where it is not defined. (This is the real reason why you usually "solve" for $\frac{d^2y}{dx^2}=0$: to find the places where it can change from one sign to the other).

Now, a fraction is undefined where the denominator is $0$; and is $0$ where the denominator is not zero and the numerator is zero.

As you note, the numerator of this second derivative is never zero. That means the second derivative is never zero. Is the second derivative ever undefined?

Once you find all the places where the second derivative could change sign, then you can figure out the sign of the second derivative everywhere else by simply evaluating. For example, say for the sake of argument that you found that the derivative is never zero, and is undefined at $-3$ and at $2$. That means that the only places where it can change signs is at $x=-3$ and at $x=2$. So, it must have one particular sign on $(-\infty,-3)$, one particular sign on $(-3,2)$, and one particular sign on $(2,\infty)$ (some of these may be the same sign, but it cannot change sign anywhere except at $x=-3$ or at $x=2$). So you could then just plug in, say, $x=-4$, $x=0$, and $x=5$ to determine the sign it has on each of those intervals, and obtain the concavity of the original function that way.

If you were to discover that the second derivative is never zero and never undefined, that would mean that it is always positive or always negative, so you could just find the value at a single point to figure out which one and so figure out which concavity the function always has.

Hope that helps.

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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.

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