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If c(x)= 5/x-2 and d(x)=x+3
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If c(x)= 5/x-2 and d(x)=x+3

If c(x)= 5/x-2 and d(x)=x+3-example-1
User BadgerPriest
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\bf \begin{cases} c(x)=\cfrac{5}{x-2}\\ d(x)=x+3\\ ----------\\ (cd)(x)\iff c(x)\cdot d(x) \end{cases}\implies \cfrac{5x+15}{x-2}

the domain, or safe values "x" can take on, are values, that do not make the fraction undefined, and that happens, when the denominator is 0

well, take a look at the denominator, what values of "x", make x-2 end up as 0?
User Thomasina
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