Question

Need Help with the composition of two rational functions. screenshot is below

Answer 1

We just have to write one function inside the other:

`(f\circ g)(x) = f(g(x))\\\\(f\circ g)(x) = f\left((13)/(x+3)\right)\\\\(f\circ g)(x) = (2)/(\left((13)/(x+3)\right)+3)\\\\(f\circ g)(x) = (2)/(~~(13+3(x+3))/(x+3)~~)\\\\(f\circ g)(x) = (2(x+3))/(13+3(x+3))\\\\\boxed{(f\circ g)(x) = (2x+6)/(3x+22)}`

Since we have a fraction with x in the denominator, we have to remove from the domain the value of that makes null the denominator. Thus,

`3x+22\\eq0\Longrightarrow 3x\\eq22\Longrightarrow x\\eq (22)/(3)`

Furthermore, we also have to remove the value of x that annuls the denominator of g(x), because that's the first function that we use. So:

`x+3\\eq0\Longrightarrow x\\eq-3`

Hence, the domain is:

`D((f\circ g)(x)) = \left]-\infty,-3\right[\cup \left]-3,(22)/(3)\right[\cup\left](22)/(3),\infty\right[`