If #f(x) = x/6 - 2# and #g(x) = 6x + 12#, how can I show #(f@g)(x) = (g@f)(x)#?
1 Answer
Jun 26, 2018
Explanation:
#"if "(f@g)(x)=(g@f)(x)=x" then"#
#f(x)" and "g(x)" are inverse functions of each other"#
#"if we show that "f(x)" and "g(x)" are inverse of each "#
#"other then the above statement should be true"#
#"let "y=x/6-2" and "y=6x+12#
#"rearrange both making x the subject"#
#6y=x-12rArrx=6y+12#
#f^-1(x)=6x+12=g(x)#
#y=6x+12rArrx=1/6(y-12)=y/6-2#
#g^-1x=x/6-2=f(x)#
#rArr(f@g)(x)=(g@f)(x)#
#color(blue)"As a check"#
#(f@g)(x)=f(6x+12)#
#color(white)(xxxxxxx)=(6x+12)/6-2=x+2-2=x#
#(g@f)(x)=g(x/6-2)#
#color(white)(xxxxxxx)=6(x/6-2)+12=x-12+12=x#
#rArr(f@g)(x)=(g@f)(x)#