Answer:
(1) [tex]g[f(x)]=\frac{8x-1}{12x-4}[/tex]
(2) [tex]g^{-1}(x)=\frac{x+1}{(3x-2)}[/tex]
(3) [tex]x=\frac{9+\sqrt{129} }{6}, \frac{9-\sqrt{129} }{6}[/tex]
Step-by-step explanation:
Given functions f(x) = (4x-1)
are [tex]g(x)=\frac{2x+1}{3x-1}[/tex]
(1) [tex]g[f(x)]=\frac{2(4x-1)+1}{3(4x-1)-1}[/tex]
[tex]=\frac{8x-2+1}{12x-3-1}[/tex]
[tex]=\frac{8x-1}{12x-4}[/tex]
(2) for [tex]g^{-1}(x)[/tex], rewrite the function g(x) in terms of an equation
[tex]y=\frac{2x+1}{3x-1}[/tex]
Substitute y in place of x and x in place of y, then solve for y.
[tex]x=\frac{2y+1}{3y-1}[/tex]
(3y-1)x = 2y+1
3xy - x = 2y + 1
3xy - 2y = x + 1
y(3x-2) = x + 1
[tex]y=(\frac{x+1}{3x-2} )[/tex]
⇒ [tex]g^{-1}(x)=\frac{x+1}{(3x-2)}[/tex]
(3) [tex]f[g(x)]=(x-1)[/tex]
[tex]f[g(x)]=4[\frac{2x+1}{3x-1}] =(x-1)[/tex]
⇒ [tex]\frac{(8x+4)-(3x-1)}{3x-1}=(x-1)[/tex]
⇒ [tex]\frac{8x-3x+4+1}{3x-1}=(x-1)[/tex]
⇒ [tex]\frac{5x+5}{(3x-1)}=(x-1)[/tex]
⇒ [tex]5x+5=(3x-1)(x-1)[/tex]
⇒ [tex]5x+5=3x^2-3x-x+1[/tex]
⇒ [tex]5x+5=3x^2-4x+1[/tex]
⇒ [tex]3x^2-9x-4=0[/tex]
⇒ [tex]x=\frac{9\pm\sqrt{(-9)^2-4(-4)\times3} }{2\times3}[/tex]
[tex]x=\frac{9\pm\sqrt{81+48} }{6}[/tex]
[tex]x=\frac{9\pm\sqrt{129} }{6}[/tex]
[tex]x=\frac{9+\sqrt{129} }{6}, \frac{9-\sqrt{129} }{6}[/tex]
