Unable to apply transformation of function $f(x)$ with $g(x)=-f(-x)$. The problem is taken from the chap. 1.1 of the book titled : Calculus Problems for the new century, by Robert Fraga.

A function $f$ has values $f(0) =3, f(2)=1$, is piecewise linear, & has the slope $-1$ if $x\lt 0$ and $1$ if $x\gt 2$. Sketch the graph of the function $g$ defined by each of the following rules.

a.$\,\,\,g(x)=f(x)$
b.$\,\,\,g(x)=-f(-x)$
c.$\,\,\,g(x)=f(x+2)$
d.$\,\,\,g(x)=f(2x)$
e.$\,\,\,g(x)=f(3x-6)$
I have prepared solutions, which are not matching in part (b) (& have confusion for part (e)) with the book's solutions, which are given in terms of graphs.
The book's solutions are shown below:

My solutions:
I assume that the curve is connected between the points $x=0$ and $x=2$. 
The equation of the curve will be given by :
Part (a): (i) $y = -x+3,\,\,\, x\le 2$
(ii) $y = x-1,\,\,\, x\gt 2$
For Part (b), my graph is wrong, as per the solution given.
Part (b): (i) $y=-(x+3),\,\,\, -(-x)\le -2$
$\implies y= -x-3,\,\,\, -x \ge 2$
$\implies y= -x-3,\,\,\, x \le -2$
(ii) $y = -(-x-1),\,\,\, -(-x)\gt -2$
$\implies y = x+1,\,\,\, -x\lt 2$
$\implies y = x+1,\,\,\, x\gt -2$
For Part (c), my graph is correct, as per the solution given; as the solution shows the ordinate axis starting from $y=1$.
Part (c): (i) $y = -(x+2)+3,\,\,\, x+2\le 2\implies y = -x+1,\,\,\, x\le 0$
(ii) $y = (x+2)-1,\,\,\, (x+2)\gt 2\implies y = x+1,\,\,\, x\gt 0$
For Part (d), my graph is correct, as per the solution given; as the solution shows the ordinate axis starting from $y=1$.
Part (d): (i) $y = -2x+3,\,\,\, 2x\le 2\implies y = -2x+3,\,\,\, x\le 1$
(ii) $y = 2x-1,\,\,\, 2x\gt 2\implies y = 2x-1,\,\,\, x\gt 1$
For Part (e), the solution given is not clear about minimum value of function being $1$; hence unsure.
Part (e): (i) $y = -(3x-6)+3= -3x+9,\,\,\, 3x-6\le 2\implies y = -3x+9,\,\,\, x\le 2\frac 23$
(ii) $y = 3x-7,\,\,\,  3x-6\gt 2\implies y = 3x-7,\,\,\,  x\gt 2\frac 23$

Edit : Have found the correct solution in part (b) by only affecting the domain with change of $x$ by $-x$ to get equations:
Part (b): (i) $y=-(x+3),\,\,\, (-x)\le 2$
$\implies y= -x-3,\,\,\, x \ge -2$
(ii) $y = -(-x-1),\,\,\, (-x)\gt 2$
$\implies y = x+1,\,\,\, x\lt -2$
But, seems like am missing theory as why for $g(x)=-f(-x)$ only the domain is affected by exchanging $x$ by $-x$; even though it means that the whole function is also negated.
 A: Perhaps you can think that the negative on the outside is changing the sign of the output, so that negative is not affecting the domain.  Also, in general you can think that if $g(x) = -f(-x)$, then that means that the graph of $g$ is the result of flipping the graph of $f$ first over the $y$-axis (that is what the inside negative is doing) and then flipping the resulting graph over the $x$-axis (that is what the outside negative is doing). 

A: Most likely you were expected to sketch this out without such details by understanding what does each operation do. That is just do what the comments did. 
part $(e)$ is fine. 
\begin{align}
f(x) &= \begin{cases} -x+3&, x\le 2 \\ x-1 &, x>2\end{cases}
\end{align}
\begin{align}
\color{red}-f(\color{blue}-x) &= \begin{cases} \color{red}-(-(\color{blue}-x)+3)&, \color{blue}-x\le 2 \\ \color{red}-(\color{blue}-x-1) &, \color{blue}-x>2\end{cases}\\
&=\begin{cases} -x-3&, x\ge -2 \\ x+1 &, x<-2\end{cases}\\
\end{align}
The negative outside just flip the graph about the $x$-axis, it doesn't affect the horizontal position.
