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Math Help - continuety of a complex function question..

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    continuety of a complex function question..

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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by transgalactic View Post
    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
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