Raymond B. answered 07/29/25
Math, microeconomics or criminal justice
show (fog)(x) = 3f(x) is an identity. check to see if it is, try x=1/2
f(1/2) = log(3/2/1/2) = log3, 3f(3) = 3log3 = log27
g(1/2) =(3x+x^2)/(3x^2+1) =(3/2+1/4)/(3/4 +1) = 7/4/7/4 = 1
f(g(1/2) = f(1) = log(2/0) which doesn't = log27
(fog)(x) = f(g(x)) = f((3x+x^2)/(3x^2+1)) = log((1+(3x+x^2)/(3x^2+1))/(1-(3x+x^2)/(3x^2+1))
3f(x) = log((1+x)^3/(1-x)^3)= 3log(1+x) - 3log(1-x)
no apparent way to show 3log(1+x) -3log(1-x) = log(1+(3x+x^2)/(3x^2+1)/(1-(3x+x^2)/3x^2+1))