[tex]\begin{gathered} (a)\text{ Solve for }f(x)=\begin{cases}3x+1\text{ if x<1} \\ 5x-1\text{ if x>1}\end{cases} \\ \text{Equate the two cases of }f(x),\text{ so that they will be continous} \\ 3x+1=5x-1 \\ 3x-5x=-1-1 \\ \frac{-2x}{-2}=\frac{-2}{-2} \\ \\ x=1 \\ \text{Plug in }x=1\text{ to both cases} \\ 3x+1\rightarrow3(1)+1\rightarrow4 \\ 5x-1\rightarrow5(1)-1\rightarrow4 \\ \\ \text{Therefore, }f(a)=4 \end{gathered}[/tex][tex]\begin{gathered} (b)\text{ Solve for }f(x)=\begin{cases}x\text{ if }x<0 \\ -x\text{ if }x>0\end{cases} \\ \\ \text{Substituting }x=0\text{ for both cases would yield }f(a)=0 \end{gathered}[/tex]
