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Let $$f(x) = \frac{x}{x-2}$$

Additionally, let $$g(x) = \frac{3}{x}$$

The domain of $f(x)$ is all reals, except for $x = 2$, and the domain for $g(x)$ is all reals, except for $x=0$.

$$(f \circ g)(x) = \frac{3}{3-2x}$$

However, the domain for this function seems to be not simply $x \ne 3/2$ but also $x \ne 0$, since it must comply with the domain of $g(x)$. However, the image of $(f\ o\ g)(x)$ does have a solution at $x = 0$, as one can also plainly tell from the function at face value. So how is this supposed to stay true? And why, in this case, need we only the domain restrictions for $g(x)$, and not say $f(x)$, saying that $x \ne 2$ as well?

Zubin Mukerjee
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sangstar
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  • Briefly speaking, you just can plug-in the values of $x$ for which both expressions are defined. Since $f(g(x))$ is well defined for $g(x) \neq 2$ and $g(x)$ is well defined for $x\neq 0$, you obtain as a consequence the above-mentioned restrictions. – user0102 Jul 23 '17 at 03:48

1 Answers1

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$$(f\circ g)(x) = \frac{\frac{3}{x}}{\frac{3}{x}-2} = \frac{x}{\color{red}{x}}\frac{3}{3-2x}$$

So, $x$ must not be $0$ i.e. $0$ is not in the domain of $x$.

Zubin Mukerjee
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Dhruv Kohli
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  • I seem to misunderstand why adding $x/x$ explains this. I can multiply, theoretically, any function by $x/x$ since it resolves to 1 and equality is maintained but it now has this domain restriction. Why does this apply here and not elsewhere? And why is what I just implied not true? – sangstar Jul 23 '17 at 03:49
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    @sangstar Multiplying by $x/x$ is not the same as multiplying by $1$! Multiplying by $x/x$ restricts the domain by excluding $0$. – Zubin Mukerjee Jul 23 '17 at 03:51
  • And why does this explain the general case for having to include the domain of $g(x)$ in this composite function? It seems to me you've proved it just by observation of the function, yet I don't see how what you've done maintains an equality if you've applied this transformation which doesn't resolve to 1. Perhaps this is true after playing with the two sides algebraically, but even so this explanation is foggy to me. – sangstar Jul 23 '17 at 03:58