For what values of x is the graph of y=xe^−4x concave down?
We have a product of functions, so we use the product rule.
$\displaystyle \frac{dy}{dx} = e^{-4x} - 4xe^{-4x} = e^{-4x}(1 - 4x)$.
$\displaystyle \frac{d^2y}{dx^2} = -4e^{-4x} - 4e^{-4x}(1 - 4x) = -4e^{-4x}(2 - 4x)$.
We require that this be greater than 0.
Notice that $\displaystyle -4e^{-4x}$ is always negative.
So for the second derivative to be positive, $\displaystyle 2 - 4x$ must also be negative (as negative times negative is positive).
$\displaystyle 2 - 4x < 0$
$\displaystyle 2 < 4x $
$\displaystyle x > \frac{1}{2}$.
So the values of x for which the graph is concave down are all those that are greater than $\displaystyle \frac{1}{2}$.