Question 1012362
gog(x) = g(g(x)).


this is no different than fog(x) = f(g(x)), when the function called f is the exact same function as the function called g.


here's an example of gog(x).


let g(x) = x^2


gog(x) = g(g(x))


since g(x) = x^2, g(g(x)) becomes g(x^2).


since g(x) = x^2, then g(x^2) = (x^2)^2 = x^4.


when the argument is x, g(x) = x^2


when the argument is x^2 rather than x, you raplace x with x^2 to get g(x^2) = (x^2)^2.


let's do it with numbers to see if that's any clearer.


g(x) = x^2


let's find g(2).


g(2) = 2^2 = 4


gog(2) = g(g(2))


since g(2) = 4, then g(g(2)) = g(4) = 4^2 = 16.


the result from the inner function is passed to the outer function and then the outer function works on that.


bottom line:


gog(x) is the same as fog(x) when f performs the exact same function as g.


if f(x) = x^2 and g(x) = x^2, you would get exactly the same result from fog(x) as you would get from gog(x).