Can someone please help me f(x)=1/3 square root 1-2x and g(x)= (x-2)/x. Find and simplify (g of g)(3)/(g of f)(-24).
The answer I got it -24/7 and i Am unsure If it is correct. Please show all work
Thank you
Too bad you didn't see fit to show your work...
f(x) = 1/3 √(1-2x)
g(x) = (x-2)/x
(g◦g)(x) = g(g(x)) = (g-2)/g
= ((x-2)/x - 2)/((x-2)/x)
= 1 - 2x/(x-2)
(g◦f)(x) = (f-2)/f
= (1/3 √(1-2x) - 2)/(1/3 √(1-2x))
= 1 - 6/√(1-2x)
(g◦g)(3) = 1 - 6/1 = -5
(g◦f)(-24) = 1 - 6/7 = 1/7
(g◦g)(3)/(g◦f)(-24) = -35
Or, using the numeric values earlier,
g(x) = (x-2)/x
g(3) = 1/3
g(g(3)) = g(1/3) = (-5/3)/(1/3) = -5
f(x) = 1/3 √(1-2x)
f(-24) = 7/3
g(7/3) = (1/3)/(7/3) = 1/7
as above