Answer:
The value(s) of a and b for which the function f(x) have a point discontinuities at x= -3 and x= 2 is;
[tex]\begin{gathered} a=\frac{6}{5} \\ b=\frac{1}{5} \end{gathered}[/tex]Explanation:
The point discontinuities for the function f(x) for x= -3 is the value of a and b for which;
[tex]2x+3=ax+3b[/tex]substitute x=-3, we have;
[tex]\begin{gathered} 2(-3)+3=a(-3)+3b \\ -6+3=-3a+3b \\ -3=-3a+3b \\ \text{divide through by 3} \\ -1=-a+b \\ -a+b=-1\ldots\ldots\ldots.i \end{gathered}[/tex]The point discontinuities for the function f(x) for x= 2 is the value of a and b for which;
[tex]ax+3b=\frac{-1}{2}x^2+3x-1[/tex]substituting x=2, we have;
[tex]\begin{gathered} ax+3b=\frac{-1}{2}x^2+3x-1 \\ a(2)+3b=\frac{-1}{2}(2)^2+3(2)-1 \\ 2a+3b=-2+6-1 \\ 2a+3b=3\ldots\ldots\ldots\text{ }ii \end{gathered}[/tex]So, we need to solve the simultaneous equation i and ii to get the value of a and b;
[tex]\begin{gathered} -a+b=-1\ldots\ldots\ldots.i \\ 2a+3b=3\ldots\ldots\ldots\text{ }ii \end{gathered}[/tex]from equation i;
[tex]b=-1+a[/tex]substituting that into equation ii;
[tex]\begin{gathered} 2a+3(-1+a)=3 \\ 2a-3+3a=3 \\ 5a-3=3 \\ 5a=3+3 \\ 5a=6 \\ a=\frac{6}{5} \end{gathered}[/tex]substitute the value of a to get b;
[tex]\begin{gathered} b=-1+a \\ b=-1+(\frac{6}{5}) \\ b=\frac{1}{5} \end{gathered}[/tex]Therefore, the value(s) of a and b for which the function f(x) have a point discontinuities at x= -3 and x= 2 is;
[tex]\begin{gathered} a=\frac{6}{5} \\ b=\frac{1}{5} \end{gathered}[/tex]