The figure shows the graph of a function g on the closed interval [-4, 4]…

(a) On each piece $g$ is a polynomial, so the only candidates are $x = -1$ and $x = 1$. At $x = -1$ the slopes are $2$ (left) and $-2$ (right). At $x = 1$ they are $-2$ (left) and $-\dfrac{2(1 - 3)}{4} = 1$ (right). Both are corners: 2 points.

(b) $F = |g|$ keeps the two corners of $g$, since $g(-1) = 3$ and $g(1) = -1$ are not zero, and $|g|$ is $g$ or $-g$ near them. New trouble can come only where $g = 0$: at $x = -\tfrac52$ and $x = \tfrac12$ (on the straight pieces) and at $x = 3$ (on the arc). At $x = -\tfrac52$ and $x = \tfrac12$ the graph crosses the axis with slopes $2$ and $-2$. At each of them $|g|$ has slope $-2$ just to the left and $2$ just to the right: two new corners. At $x = 3$ the arc only touches the axis: $g(x) = -\dfrac{(x - 3)^2}{4} \le 0$, so near $x = 3$, $F(x) = \dfrac{(x - 3)^2}{4}$, a smooth parabola with slope 0. So $F$ fails at $-1$, $1$, $-\tfrac52$ and $\tfrac12$: 4 points.

(c) For $x \ge 0$, $H(x) = g(x)$, and for $x < 0$, $H(x) = g(-x)$. So $H$ has the corner of $g$ at $x = 1$ and its mirror image at $x = -1$. At $x = 0$: near 0, $H(x) = 1 - 2|x|$, with slopes $2$ on the left and $-2$ on the right, another corner. The point $x = -1$ of $g$ itself plays no part, because $H$ uses $g$ only on $[0, 4)$. So $H$ fails at $-1$, $0$ and $1$: 3 points.

(d) At $x = 2$, $g(2) = -\tfrac14 < 0$, so near $x = 2$, $F(x) = -g(x) = \dfrac{(x - 3)^2}{4}$ and $F'(x) = \dfrac{x - 3}{2}$. So $F'(2) = -0.50$.

(e) $F(x) = 1$ means $g(x) = 1$ or $g(x) = -1$. On $[-4, -1]$: $2x + 5 = 1$ gives $x = -2$, and $2x + 5 = -1$ gives $x = -3$. On $(-1, 1]$: $1 - 2x = 1$ gives $x = 0$, and $1 - 2x = -1$ gives $x = 1$. On $(1, 4]$ the arc runs from just above $-1$ up to $0$ and back down to $-\tfrac14$, so it never reaches $\pm 1$ again ($-\tfrac{(x - 3)^2}{4} = -1$ only at $x = 1$ and $x = 5$). The solutions are $-3, -2, 0, 1$: 4. The one at $x = 1$ is easy to miss, because $F$ only touches the level 1 there and does not cross it.