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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: \[f(x) = \sqrt{x  1} \\ g(x) = 3x + 1\]
\[P(x) = \left(\dfrac{b}{g}\right)(x)\]
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have so far: \[P(x) = \left(\dfrac{\sqrt{x  1}}{3x + 1}\right)(x)\] I also know the deniominator \(\ne\) \(0\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just noticed the numerator cannot be 1, or smaller as that would cause imaginary numbers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, it can be \(1\) since that just makes the numerator zero.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2you need the following to be true \(x1\ge 0\) 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 \[\{x\mid x\ge 1\}\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2that should say \(x\ne \frac{1}{3}\) not \(x\ge \frac{1}{3}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All 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.))

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Nah, 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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