So I stumbled upon this problem while exercising and I faced some difficulties:
I managed to answer it correctly, D; however, it took a really long time. So it got me wondering if I might be approaching the problem the wrong way. so what is the easiest way to answer this problem? Thank you for your help!
A differentiable function $f(x)$ defined on an interval is concave if (and only if) its first derivative is decreasing over the interval. You are given $f^\prime (x)$ and you can check whether it is decreasing by studying where the second derivative $$f^{\prime\prime} (x) = (4x^3-4x) e^{x^2-2x^2+1} \le 0$$ Since the sign of $f^{\prime\prime} (x)$ only depends on the sign $(4x^3-4x) = 4x(x^2-1)$, you only need to solve for $4x(x^2-1) \le 0$ in your given interval. Doing so reveals that the right answer is D.