# Thread: continuety of a complex function question..

1. ## continuety of a complex function question..

2. Originally Posted by transgalactic
What do you mean by complex function? I assme you mean that a difficult function...I assume this because the sign function is only defined on the reals.

1. $f(g(x))=1+\text{sign}(x)^2$

Note that $\text{sign}(x)=\left\{\begin{array}{rcl} -1 & \mbox{if} & x<0 \\ 0 & \mbox{if} & x=0\\ 1 & \mbox{if} & x>0 \end{array} \right.$

So $1+\text{sign}(x)^2=\left\{ \begin{array}{rcl} 2 & \mbox{if} & x\ne0\\ 1 & \mbox{if} & x=0 \end{array} \right.$

So $f(g(x))$ is continuous everywhere except zero.

$g(f(x))=\text{sign}\left(1+x^2\right)$. Now since $1+x^2>0\implies \text{sign}\left(x^2+1\right)=1~~\forall x \in\mathbb{R}$

2. $f(g(x))=\sin(\ln(x))$

Now it is known that the composition of two continuous functions is continous. So at all points of definition this function is continuous so it is continuous for all values of $x>0$

$g(f(x))=\ln(\sin(x))$

Same concept here except we have that the function is defined on the set $S=\left\{x:\sin(x)>0\right\}$

3. $f(g(x))=\lfloor \text{sign}(x)\rfloor$

Notice that this changes nothing since $\text{sign}(x)\in\mathbb{Z}$

$g(f(x))=\text{sign}\left(\lfloor x\rfloor\right)$

Note that $x<0\implies \lfloor x\rfloor<0$

$0\leqslant x<1\implies \lfloor x\rfloor=0$

$x\geqslant 1\implies \lfloor x\rfloor>0$