Question

Precalculus please help!!!!

Answer 1

1)

f(x) = $\frac{2x + 6}{3x}$         g(x) = $\frac{\sqrt{x} }{3x}$

f(x) - g(x) = $\frac{2x + 6}{3x}$ -  $\frac{\sqrt{x} }{3x}$

              =  $\frac{2x + 6 - \sqrt{x} }{3x}$

2)

f(x) = $\frac{3x - 1}{x - 4}$            g(x) = $\frac{x + 1}{x}$

f(g(x)) = $\frac{3(\frac{x + 1}{x}) - 1}{(\frac{x + 1}{x}) - 4}$

        = $\frac{2x + 3}{-3x + 1}$

DOMAIN: x ≠ {4, 0, $\frac{1}{3}$ } ⇒ (-∞, $\frac{1}{3}$ ) U ($\frac{1}{3}$ , 0) U (0, 4) U (4, ∞)

RANGE: y ≠ {$\frac{-2}{3}$ }   ⇒  (-∞, $\frac{-2}{3}$ ) U ($\frac{-2}{3}$,  ∞)

Answer 2

Answer:

$f(x)=2x+\frac{6}{3x}\\\\g(x)=\sqrt{x}-\frac{8}{3x}\\\\(f-g)x=f(x)-g(x)\\\\=2x+\frac{6}{3x}-\sqrt{x}+\frac{8}{3x}\\\\=2x-\sqrt{x}+\frac{14}{3x}\\\\2.\rightarrow f(x)=3x-\frac{1}{x-4}\\\\g(x)=x+\frac{1}{x}\\\\fog(x)=f(g(x))\\\\=f(x+\frac{1}{x})\\\\=3\times(x+\frac{1}{x})-\frac{1}{x+\frac{1}{x}-4}\\\\fog(x)=3\times\frac{(x^2+1)}{x}-\frac{x}{x^2-4x+1}$

Domain of gof(x) will be

  x≠0---

x²-4x+1≠0

$x\neq \frac{4\pm \sqrt{16-4}}{2}\\\\x\neq \frac{4\pm \sqrt{12}}{2}\\\\x\neq 2 \pm 2 \sqrt{3}$

x=R-($x\neq \frac{4\pm \sqrt{16-4}}{2}\\\\x\neq \frac{4\pm \sqrt{12}}{2}\\\\x\neq 2 \pm 2 \sqrt{3}$,0)