Calculus 10th Edition

The function is continuous when $a=-1,$ and $b=1.$
$\lim\limits_{x\to-1^-}f(x)=2.$ $\lim\limits_{x\to-1^+}f(x)=a(-1^+)+b=b-a.$ $\lim\limits_{x\to3^-}f(x)=a(3^-)+b=3a+b.$ $\lim\limits_{x\to3^+}f(x)=-2.$ For the function to be continuous, both $\lim\limits_{x\to-1^-}f(x)=\lim\limits_{x\to-1^+}f(x)$ and $\lim\limits_{x\to3^-}f(x)=\lim\limits_{x\to3^+}f(x)\to$ $b-a=2\to b=a+2.$ $3a+b=-2.$ Solving the system of equations gives us $3a+(a+2)=-2\to4a=-4\to a=-1$ and $b=(-1)+2=1.$