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# Let $f : R \to R$ be the Signum Function defined as $f(x) = \left \{ \begin {array} {1 1} 1, & \quad \text { x > 0} \\ 0, & \quad \text { x =0} \\-1, & \quad \text { x <0} \\ \end {array} \right.$ and $g:R \to R$ be the greatest Integer Function given by $g(x)=[x]$ where $[x]$ is a greatest integer less thar or equal to $x$ Then, does $fog$ and $gof$ coincide in $(0,1]$?.

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Toolbox:
• Given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
Given a Signum function $f:R \to R$ defined by $f(x)= \left\{ \begin{array}{1 1} 1 & \quad x > 0 \\ 0 & \quad x=0 \\ -1 & \quad x<0 \end{array} \right.$ and the greatest integer function $g:R \to R$ defined by $g(x)=[x]$ greatest integer less than or equal to x, where $x \in (0,1]$.
$\textbf {Step 1: Calculating gof}$
$x \in (0,1] \Rightarrow f(x) = 1$ as $x>0$.
Therefore, $gof = g(f(x)) = g(1) = [1] = 1$.
$\textbf {Step 2: Calculating fog}$
$x \in (0,1] \Rightarrow g(x) = [1] = 1$ if $x=0$ or $g(x) = [0]$ if $x \in (0,1)$
$\Rightarrow fog = f(g(x) = \left\{ \begin{array}{1 1} f(1) & \quad x =1 \\ f(0) & \quad x\in(0,1) \end{array} \right. = \left\{ \begin{array}{1 1} 1 & \quad x =1 \\ 0 & \quad x\in(0,1) \end{array} \right.$
Thus $x \in (0,1]$ we have $fog(x)=0$ and $\; gof(x)=1$
Therefore, they do not coincide.
answered Feb 28, 2013 by
edited Mar 20, 2013