Question

Let $f:R\to R$ and $g:R\to R$ be defined as
$f(x)=\{\begin{array}{ll}x+a,& x<0\\ \mid x-1l,& x\ge 0\end{array}.$ and
$g(x)=\{\begin{array}{ll}x+1,& x<0\\ (x-1{)}^2+b,& x\ge 0\end{array}.$
where a, b are non-negative real numbers. If (gof) $(x)$ is continuous for all $x\in R$, then $a+b$ is equal to ________.

Answer 1

Answer: 1

$g[f(x)]=\left[\begin{array}{cc}f(x)+1& f(x)<0\\ (f(x)-1{)}^2+b& f(x)\ge 0\end{array}\right]$
$g[f(x)]=\left[\begin{array}{cc}x+a+1& x+a<0\&x<0\\ |x-1|+1& |x-1|<0\&x\ge 0\\ (x+a-1{)}^2+b& x+a\ge 0\&x<0\\ (|x-1|-1{)}^2+b& |x-1|\ge 0\&x\ge 0\end{array}\right]$
$g[f(x)]=\left[\begin{array}{cc}x+a+1& x\in (-\infty ,-a)\&x\in (-\infty ,0)\\ |x-1|+1& x\in \varphi \\ (x+a-1{)}^2+b& x\in [-a,\infty )\&x\in (-\infty ,0)\\ (|x-1|-1{)}^2+b& x\in R\&x\in [0,\infty )\end{array}\right]$
$g[f(x)]=\left[\begin{array}{cc}x+a+1& x\in (-\infty ,-a)\\ (x+a-1{)}^2+b& x\in [-a,0)\\ (|x-1|-1{)}^2+b& x\in [0,\infty )\end{array}\right]$
$g(f(x))$ is continuous
at $x=-a\&$at $x=0$
$1=b+1\&(a-1{)}^2+b=b$
$b=0\&a=1$
$\Rightarrow a+b=1$