On combining functions, and composite functions. How do I deriving, the domain, and range of such functions.

An example: $$f(x) = \sqrt{x - 1} \ g(x) = 3x + 1$$
$$P(x) = \left(\dfrac{b}{g}\right)(x)$$

Question

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 · Answer

I have so far: $$P(x) = \left(\dfrac{\sqrt{x - 1}}{3x + 1}\right)(x)$$

I also know the deniominator $\ne$ $0$.

anonymous · Answer

I just noticed the numerator cannot be 1, or smaller as that would cause imaginary numbers.

anonymous · Answer

Wait, it can be $1$ since that just makes the numerator zero.

zzr0ck3r · Answer

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 $$\{x\mid x\ge 1\}$$

zzr0ck3r · Answer

that should say $x\ne \frac{-1}{3}$ not $x\ge \frac{1}{3}$

anonymous · Answer

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.))

zzr0ck3r · Answer

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