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# Let $f,\, g\, and\, h$ be functions from $R\, to\, R.$ Show that $(f+g) oh = foh + goh$

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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 $gof (x)=g(f(x))\;for\; all \;x \in A$
We have to prove: L.H.S.: $(f+g) oh =$ R.H.S.: $foh + goh$.
Consider L.H.S. $(f+g) oh$,
$\Rightarrow (f+g) oh = (f+g) o (h(x))$
$\Rightarrow (f+g) oh = f(h(x))+ g(h(x))$
We know that given two functions $f:A \to B$ and $g:B \to C$, then composition of $f$ and $g$, $gof:A \to C$ by $gof (x)=g(f(x))\;for\; all \;x \in A$
$\Rightarrow f(h(x)) = foh$ and $g(h(x)) = goh$
$\Rightarrow$ L.H.S. $= f(h(x)) + g(h(x)) = foh + goh$ = R.H.S.
answered Feb 25, 2013 by
edited Mar 19, 2013

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