Let f(x)=1/x and g(x)= x^2-2x. What two numbers are not in the domain of fog?

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Let f(x)=1/x and g(x)= x^2-2x. What two numbers are not in the domain of fog?

Mathematics
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Clearly g(x) = x^2-2x has no restrictions, so simply find the restrictions of f(g(x)) = f(x^2-2x) = 1/(x^2-2x)
\[f(g(x))=\dfrac{1}{x^2-2x}\] what values of \(x\) make that expression go crazy ?
Well, is 0 one of them? @ganeshie8

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Yes, simply set the bottom equal to 0 and solve x : \[x^2-2x=0\]
@ganeshie8 x= -2? so it would be 0,-4?
-2*
\(x^2-2x=0 \) \(x(x-2)=0\) \(x=0\) or \(x-2=0\) that gives you \(x~=~0, ~~2\)
\(0, 2\) are the two numbers that must be excluded from the domain of fog
Usually when I set it up for the domain I write it as \[x^2-2x \neq 0\] then solve for x, then you know you are looking for the restrictions. That's just a quick tip for rational functions :P

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