Question

Precalculus please help!!!!

1) Find (f-g)(x) when f(x)=2x+6/3x and g(x)=(sqrt)x-8/3x

2)Determine the domain and function of (fog)(x) where f(x)=3x-1/x-4 and g(x)=x+1/x

Answer 1

`f(x)=2x+(6)/(3x)\\\\g(x)=√(x)-(8)/(3x)\\\\(f-g)x=f(x)-g(x)\\\\=2x+(6)/(3x)-√(x)+(8)/(3x)\\\\=2x-√(x)+(14)/(3x)\\\\2.\rightarrow f(x)=3x-(1)/(x-4)\\\\g(x)=x+(1)/(x)\\\\fog(x)=f(g(x))\\\\=f(x+(1)/(x))\\\\=3*(x+(1)/(x))-(1)/(x+(1)/(x)-4)\\\\fog(x)=3*((x^2+1))/(x)-(x)/(x^2-4x+1)`

Domain of gof(x) will be

x≠0---

x²-4x+1≠0

`x\\eq (4\pm √(16-4))/(2)\\\\x\\eq (4\pm √(12))/(2)\\\\x\\eq 2 \pm 2 √(3)`

x=R-(
`x\\eq (4\pm √(16-4))/(2)\\\\x\\eq (4\pm √(12))/(2)\\\\x\\eq 2 \pm 2 √(3)`,0)

Answer 2

1)

f(x) =
`(2x + 6)/(3x)` g(x) =
`(√(x) )/(3x)`

f(x) - g(x) =
`(2x + 6)/(3x)` -
`(√(x) )/(3x)`

=
`(2x + 6 - √(x) )/(3x)`

2)

f(x) =
`(3x - 1)/(x - 4)` g(x) =
`(x + 1)/(x)`

f(g(x)) =
`(3((x + 1)/(x)) - 1)/(((x + 1)/(x)) - 4)`

=
`(2x + 3)/(-3x + 1)`

DOMAIN: x ≠ {4, 0,
`(1)/(3)` } ⇒ (-∞,
`(1)/(3)` ) U (
`(1)/(3)` , 0) U (0, 4) U (4, ∞)

RANGE: y ≠ {
`(-2)/(3)` } ⇒ (-∞,
`(-2)/(3)` ) U (
`(-2)/(3)`, ∞)