For 14, I think the definition of the absolute value will make things easier. Recall that
[tex]|x|=\begin{cases}x&\text{for }x\ge0\\-x&\text{for }x<0\end{cases}[/tex]
So we have
[tex]|x-3|=\begin{cases}x-3&\text{for }x\ge3\\3-x&\text{for }x<3\end{cases}[/tex]
[tex]|x+1|=\begin{cases}x+1&\text{for }x\ge-1\\-x-1&\text{for }x<-1\end{cases}[/tex]
Putting these together, we have 3 different cases to consider:
[tex]f(x)=\begin{cases}(3-x)-2(-x-1)=x+5&\text{for }x<-1\\(3-x)-2(x+1)=1-3x&\text{for }-1\le x<3\\(x-3)-2(x+1)=-x-5&\text{for }x\ge3\end{cases}[/tex]
Then we check the derivative, noting that we shouldn't expect the derivative to be continuous at [tex]x=-1[/tex] and [tex]x=3[/tex], so we ignore those exact cases:
[tex]f'(x)=\begin{cases}1&\text{for }x<-1\\-3&\text{for }-1<x<3\\-1&\text{for }x>3\end{cases}[/tex]
This tells us that [tex]f(x)[/tex] is increasing on [tex](-2,-1)[/tex] and decreasing on [tex](-1,3)[/tex]. We know that [tex]f(-2)=3[/tex] and [tex]f(3)=-8[/tex], but if we can verify that [tex]f(x)[/tex] is continuous at [tex]x=-1[/tex], then we can use the trends from the derivative test above to show there's at least a local maximum at that point.
We have [tex]f(-1)=4[/tex], while
[tex]\displaystyle\lim_{x\to-1^-}f(x)=\lim_{x\to-1}x+5=4[/tex]
[tex]\displaystyle\lim_{x\to-1^+}f(x)=\lim_{x\to-1}1-3x=4[/tex]
so [tex]f(x)[/tex] is indeed continuous. So, [tex]f(-1)=4[/tex] is an absolute maximum and [tex]f(3)=-8[/tex] is an absolute minimum.
For 15, the fact that [tex]f[/tex] is differentiable and attains an extreme value at [tex]x=2[/tex] means that [tex]f'(2)[/tex] exists, and that [tex]f'(2)=0[/tex]. Then for part (a),
[tex]g(2)=2f(2)+1=1[/tex]
[tex]h(2)=2f(2)+2+1=3[/tex]
[tex]g'(2)=f(2)+2f'(2)=0[/tex]
[tex]h'(2)=f(2)+2f'(2)+1=1[/tex]
For part (b), in order for [tex]g[/tex] or [tex]h[/tex] to have extreme values at [tex]x=2[/tex], we would need to have [tex]g'(2)=0[/tex] (which is true, as shown above) and [tex]h'(2)=0[/tex] (which is not).
