Question

f(x)= 2x + 1/ x-3 , g(x) = 4-2x/ x+2 (Fog)(x) = (gof)(x) =​

Answer 1

Materi : Fungsi dan Relasi

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

g(x) = ( - 2x + 4 )/( x + 2 )

( f o g )( x )

= f( [ -2x + 4 ]/[ x + 2 ] )

= { 2( [ -2x + 4 ]/[ x + 2 ] ) + 1 }/{ [ -2x + 4 ]/[ x + 2 ] - 3 }

= { [ - 4x + 8 ]/[ x + 2 ] + [ x + 2 ]/[ x + 2 ] }/{ [ - 2x + 4 ]/[ x + 2 ] - [ 3x + 6 ]/[ x + 2 ] }

= { [ - 3x + 10 ]/[ x + 2 ] }/{ [ - 5x - 2 ]/[ x + 2 ] )

= ( - 3x + 10 )/( - 5x - 2 )

( g o f )( x )

= g( [ 2x + 1 ]/[ x - 3 ] )

= { -( [ 2x + 1 ]/[ x - 3 ] ) + 4 }/{ [ 2x + 1 ]/[ x - 3 ] + 2 }

= { [ - 2x - 1 ]/[ x - 3 ] + [ 4x - 12 ]/[ x - 3 ] }/{ [ 2x + 1 ]/[ x - 3 ] + [ 2x - 6 ]/[ x - 3 ] }

= { [ 2x - 13 ]/[ x - 3 ] }/{ [ 4x - 5 ]/[ x - 3 ] }

= ( 2x - 13 )/( 4x - 5 )

Semoga bisa membantu

[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]

Answer 2

[tex]\begin{aligned}\bullet\ &(f\circ g)(x)=\frac{-3x+10}{-5x-2}\\\bullet\ &(g\circ f)(x)\vphantom{\Bigg|}=-\frac{14}{4x-5}\end{aligned}[/tex]
 

Penjelasan dengan langkah-langkah

Komposisi Fungsi

Diberikan:

[tex]\begin{aligned}f(x)=\frac{2x+1}{x-3}\,,\ g(x)=\frac{4-2x}{x+2}\end{aligned}[/tex]

Maka:

[tex]\begin{aligned}(f\circ g)(x)&=f\left(g(x)\right)\\&=\frac{2g(x)+1}{g(x)-3}\\&=\frac{2\left(\dfrac{4-2x}{x+2}\right)+1}{\left(\dfrac{4-2x}{x+2}-3\right)}\\&=\frac{\left(\dfrac{8-4x+x+2}{\cancel{x+2}}\right)}{\left(\dfrac{4-2x-3(x+2)}{\cancel{x+2}}\right)}\\&=\frac{8-4x+x+2}{4-2x-3x-6)}\\(f\circ g)(x)&\vphantom{\Bigg|}=\frac{-3x+10}{-5x-2}\\\end{aligned}[/tex]

[tex]\begin{aligned}(g\circ f)(x)&=g\left(f(x)\right)\\&=\frac{4-2f(x)}{f(x)+2}\\&=\frac{4-2\left(\dfrac{2x+1}{x-3}\right)}{\left(\dfrac{2x+1}{x-3}+2\right)}\\&=\frac{\left(\dfrac{4(x-3)-2(2x+1)}{\cancel{x-3}}\right)}{\left(\dfrac{2x+1+2(x-3)}{\cancel{x-3}}\right)}\\&=\frac{4(x-3)-2(2x+1)}{2x+1+2(x-3)}\\&=\frac{\cancel{4x}-12-\cancel{4x}-2}{2x+1+2x-6}\\(g\circ f)(x)&\vphantom{\Bigg|}=-\frac{14}{4x-5}\end{aligned}[/tex]