Quick question regarding Analytical Applications of Differentiation

When $-3 < x < -1$, the function is above the $x$-axis, and decreasing linearly. Thus $g(x) > 0$, $g'(x) < 0$, and $g''(x) = 0$.

When $-1 < x < 2$, the function is above the $x$-axis, and increasing more and more slowly (it is concave down). Thus $g(x) > 0$, $g'(x) > 0$, and $g''(x) < 0$.

When $2 < x < 5$, the function is above the $x$-axis, and decreasing more and more quickly (it is concave down). Thus $g(x) > 0$, $g'(x) < 0$, and $g''(x) < 0$.

When $5 < x < 6$, the function is above the $x$-axis, and increasing linearly. Thus $g(x) > 0$, $g'(x) > 0$, and $g''(x) = 0$.

The critical points of $f$ occur when $f'(x) = 0$. This means $e^{-x}(1-x) = 0$. Multiplying both sides by $e^x$ gives that $x = 1$. To determine if this is a relative maximum or minimum, we look at the second derivative (product rule) to get $f''(x) = -e^{-x}(1-x) -e^{-x}$. At $x = 1$ we get $f''(1) = e^{-1}(1-1) - e^{-1} = -1/e < 0$. Thus this is a relative maximum.

$f$ has an inflection point whenever the second derivative changes sign. We have $f''(x) = -e^{-x}(1-x) -e^{x} = e^{-x}(x-2)$. This changes sign at $x=2$, since $e^{-x}$ is always positive.

We need to take the derivative of $h$. It is $h'(1) = 2(f'(1)g(1) + f(1)g'(1))$. Using the table, we have that $g'(1) > 0$. Furthermore, we already know that $f(1) = 1/e > 0$ and $f'(1) = 0$. That means $h'(1) = 2f(1)g'(1) > 0$. So the function is increasing at $x=1$.