Question

HELP ASAP PLZ i need help !help me :((

Answer 1

Answer:

(1) $g[f(x)]=\frac{8x-1}{12x-4}$

(2)  $g^{-1}(x)=\frac{x+1}{(3x-2)}$

(3)    $x=\frac{9+\sqrt{129} }{6}, \frac{9-\sqrt{129} }{6}$

Step-by-step explanation:

Given functions f(x) = (4x-1)

are                     $g(x)=\frac{2x+1}{3x-1}$

(1)  $g[f(x)]=\frac{2(4x-1)+1}{3(4x-1)-1}$

                $=\frac{8x-2+1}{12x-3-1}$

                $=\frac{8x-1}{12x-4}$

(2) for $g^{-1}(x)$, rewrite the function g(x) in terms of an equation

$y=\frac{2x+1}{3x-1}$

Substitute y in place of x and x in place of y, then solve for y.

$x=\frac{2y+1}{3y-1}$

(3y-1)x = 2y+1

3xy - x = 2y + 1

3xy - 2y = x + 1

y(3x-2) = x + 1

$y=(\frac{x+1}{3x-2} )$

⇒ $g^{-1}(x)=\frac{x+1}{(3x-2)}$

(3)    $f[g(x)]=(x-1)$

       $f[g(x)]=4[\frac{2x+1}{3x-1}] =(x-1)$

⇒    $\frac{(8x+4)-(3x-1)}{3x-1}=(x-1)$

⇒    $\frac{8x-3x+4+1}{3x-1}=(x-1)$

⇒    $\frac{5x+5}{(3x-1)}=(x-1)$

⇒    $5x+5=(3x-1)(x-1)$

⇒    $5x+5=3x^2-3x-x+1$

⇒   $5x+5=3x^2-4x+1$

⇒   $3x^2-9x-4=0$

⇒    $x=\frac{9\pm\sqrt{(-9)^2-4(-4)\times3} }{2\times3}$

   $x=\frac{9\pm\sqrt{81+48} }{6}$

   $x=\frac{9\pm\sqrt{129} }{6}$

   $x=\frac{9+\sqrt{129} }{6}, \frac{9-\sqrt{129} }{6}$