Finding critical points of an exponential function

Let $u(x) = x^2 - 4x$, so $g(x) = e^{u(x)}$. By the chain rule, $g'(x) = e^{u(x)} \cdot u'(x)$. Since $u'(x) = 2x - 4$, we get $g'(x) = e^{x^2 - 4x}(2x - 4)$.

To find critical points, set $g'(x) = 0$. Because $e^{x^2-4x} > 0$ for all $x$, this factor can never be zero, so we need $2x - 4 = 0$, giving $x = 2$. Since $g'(x)$ is defined everywhere, the only critical point is at $x = 2$.

No. The exponential factor e raised to the power of x squared minus four x is always strictly positive for every real input, since exponential functions never output zero. This is exactly why, in part a, the entire task of solving g prime of x equals zero reduced to solving only the polynomial factor two x minus four equals zero. Recognizing that the exponential piece can be ignored when hunting for zeros is the key shortcut used there.