Question

Algebra 2 Functions operations question.

Answer 1

B.
`\left((f)/(g)\right)(x)=(1)/(x-2)` domain is
`x\in \left(-\infty, -(5)/(3)\right)\cup \left(-(5)/(3),2\right)\cup (2,\infty).`

Explanation:

If
`f(x)=3x+5` and
`g(x)=3x^2-x-10,` then

`\left((f)/(g)\right)(x)=(3x+5)/(3x^2-x-10)=(3x+5)/(3x^2+5x-6x-10)=(3x+5)/(x(3x+5)-2(3x+5))=(3x+5)/((3x+5)(x-2)).`

Note that the denominator cannot be equal to 0, so

`3x+5\\eq 0 \text{ and } x-2\\eq0\\ \\x\\eq-(5)/(3)\text{ and }x\\eq 2.`

If
`x\in \left(-\infty, -(5)/(3)\right)\cup \left(-(5)/(3),2\right)\cup (2,\infty),` then

`\left((f)/(g)\right)(x)=(3x+5)/((3x+5)(x-2))=(1)/(x-2).`