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anonymous

  • one year ago

On combining functions, and composite functions. How do I deriving, the domain, and range of such functions. An example: \[f(x) = \sqrt{x - 1} \\ g(x) = 3x + 1\] \[P(x) = \left(\dfrac{b}{g}\right)(x)\]

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  1. anonymous
    • one year ago
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    I have so far: \[P(x) = \left(\dfrac{\sqrt{x - 1}}{3x + 1}\right)(x)\] I also know the deniominator \(\ne\) \(0\).

  2. anonymous
    • one year ago
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    I just noticed the numerator cannot be 1, or smaller as that would cause imaginary numbers.

  3. anonymous
    • one year ago
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    Wait, it can be \(1\) since that just makes the numerator zero.

  4. zzr0ck3r
    • one year ago
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    you need the following to be true \(x-1\ge 0\) and \(3x+1\ne 0\) For the first you get \(x\ge 1\) and for the second you get \(x>\frac{-1}{3}\) The \(x\ge 1\) trumps the \(x\ge \frac{_1}{3}\) so the domain is \[\{x\mid x\ge 1\}\]

  5. zzr0ck3r
    • one year ago
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    that should say \(x\ne \frac{-1}{3}\) not \(x\ge \frac{1}{3}\)

  6. anonymous
    • one year ago
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    All right, thank you. (Sorry for the late reply, I was away from my keyboard.) Sub question: Would \(\{x \in ℝ\mid x\ge 1\}\) be the same as you wrote? (That is the notation use in my class on a question like this, so I wonder if it is just an optional notation, or has a significant meaning in this case. (I understand that is means x is all reals, just that does your notation already imply that without the usage of it, or am I using it wrong in this case.))

  7. zzr0ck3r
    • one year ago
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    Nah, I was just assuming \(\mathbb{R}) was the universal set, in which case it is implied. But I should be more accurate.

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