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anonymous

  • one year ago

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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  1. ganeshie8
    • one year ago
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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)

  2. ganeshie8
    • one year ago
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    \[f(g(x))=\dfrac{1}{x^2-2x}\] what values of \(x\) make that expression go crazy ?

  3. anonymous
    • one year ago
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    Well, is 0 one of them? @ganeshie8

  4. ganeshie8
    • one year ago
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    Yes, simply set the bottom equal to 0 and solve x : \[x^2-2x=0\]

  5. anonymous
    • one year ago
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    @ganeshie8 x= -2? so it would be 0,-4?

  6. anonymous
    • one year ago
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    -2*

  7. ganeshie8
    • one year ago
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    \(x^2-2x=0 \) \(x(x-2)=0\) \(x=0\) or \(x-2=0\) that gives you \(x~=~0, ~~2\)

  8. ganeshie8
    • one year ago
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    \(0, 2\) are the two numbers that must be excluded from the domain of fog

  9. Astrophysics
    • one year ago
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    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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