+36916 f(g(x)) put in g(x) which is 9-x in for 'x' in the definition of f(x)
(1-x)^2 put in 9-x for 'x'
(1 - (9-x)) ^2
= (x -8)^2 = x^2 -16x+64
You should be able to do the other one now...... 
+115 thanks. I am getting confused on the second one because in the problem you did, you subtracted the 1-9. In the second problem I tried doing that same thing. It was incorrect. 9-(1-x)^2= (8-x)^2 =x^2+16x+64 that was incorrect. I guess Im wondering where I went wrong?
+118608
Hi Shaezy,
It is good to see you interacting here.
Your substitution is perfect, EP taught you well, it is your simplifying that is incorrect.
\(If \;\;f(x) = (1 − x)^2 \qquad g(x) = 9 − x,\\ find\;\; f(g(x)) \;\;and ;\;g(f(x)). \\\\~\\ g(f(x))= 9-(1-x)^2\)
So far so good. BUT
\(\text{This is NOT }(8-x)^2\\ \text{it is } \quad9-[(1-x)^2] \)
You have to expand the bracket and then take away ALL the terms from 9.
Have another go at it 
