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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

Respuesta :

tramserran

1)

f(x) = [tex]\frac{2x + 6}{3x}[/tex]         g(x) = [tex]\frac{\sqrt{x} }{3x}[/tex]

f(x) - g(x) = [tex]\frac{2x + 6}{3x}[/tex] -  [tex]\frac{\sqrt{x} }{3x}[/tex]

              =  [tex]\frac{2x + 6 - \sqrt{x} }{3x}[/tex]

2)

f(x) = [tex]\frac{3x - 1}{x - 4}[/tex]            g(x) = [tex]\frac{x + 1}{x}[/tex]

f(g(x)) = [tex]\frac{3(\frac{x + 1}{x}) - 1}{(\frac{x + 1}{x}) - 4}[/tex]

        = [tex]\frac{2x + 3}{-3x + 1}[/tex]

DOMAIN: x ≠ {4, 0, [tex]\frac{1}{3}[/tex] } ⇒ (-∞, [tex]\frac{1}{3}[/tex] ) U ([tex]\frac{1}{3}[/tex] , 0) U (0, 4) U (4, ∞)

RANGE: y ≠ {[tex]\frac{-2}{3}[/tex] }   ⇒  (-∞, [tex]\frac{-2}{3}[/tex] ) U ([tex]\frac{-2}{3}[/tex],  ∞)

Answer:

[tex]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}[/tex]

Domain of gof(x) will be

  x≠0---

x²-4x+1≠0

[tex]x\neq \frac{4\pm \sqrt{16-4}}{2}\\\\x\neq \frac{4\pm \sqrt{12}}{2}\\\\x\neq 2 \pm 2 \sqrt{3}[/tex]

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