Let f:RR and g:RR be defined as f(x)=casesx+a, & x

Question

Let $f:\mathbb{R}\to\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ be defined as $f(x)=\begin{cases}x+a, & x<0\|x-1|, & x\ge0\end{cases}$ and $g(x)=\begin{cases}x+1, & x<0\(x-1)^2+b, & x\ge0\end{cases}$, where $a,b$ are non-negative real numbers. If $(g\circ f)(x)$ is continuous for all $x\in\mathbb{R}$, then $a+b$ is equal to..........

Answer: 1

Accepted Answer

$$g[f(x)]=\begin{cases}x+a+1, & x<-a\(x+a-1)^2+b, & -a\le x<0\(|x-1|-1)^2+b, & x\ge0\end{cases}$$
Continuity at $x=-a$: $1=1+b\Rightarrow b=0$.

Continuity at $x=0$: $(a-1)^2+b=b\Rightarrow(a-1)^2=0\Rightarrow a=1$.

So $a+b=1$.

Solutions · 1