If f(x)=casesx+a, & x0|x-4|, & x0cases and g(x)=casesx+1, & x

Question

If $f(x)=\begin{cases}x+a, & x\le0\|x-4|, & x>0\end{cases}$ and $g(x)=\begin{cases}x+1, & x<0\(x-4)^2+b, & x\ge0\end{cases}$ are continuous on $\mathbb{R}$, then $(g\circ f)(2)+(f\circ g)(-2)$ is equal to:

Accepted Answer

Continuity of $f$ at $0$: $a=|{-4}|=4$. Continuity of $g$ at $0$: $1=(0-4)^2+b\Rightarrow b=1-16=-15$.

$(g\circ f)(2)=g(f(2))=g(|2-4|)=g(2)=(2-4)^2-15=4-15=-11$.

$(f\circ g)(-2)=f(g(-2))=f(-2+1)=f(-1)=-1+a=-1+4=3$.

Sum $=-11+3=-8$.

Solutions · 1