Let us understand what we need to do for (fcircg)(x) and (gcircf)(x). Before moving ahead let us understand evaluating a function.
Example: Evaluating a function
f(x) = x^2 + 1
Let us evaluate this function at x=2
f(2) = (2)^2+1color(red)" Note here " x color(red)" is replaced by " 2
f(2) =4 + 1
f(2) = 5
If we have to evaluate this function at x=a then
f(a) = (a)^2+1 color(red)" Note here " x color(red)" is replaced by " a
If we have to replace it with a function g(x) then
f(g(x)) = (g(x))^2 + 1
We can see for evaluating we just plug in place of x
f(g(x)) is same as (fcircg)(x)
(fcircg)(x) = f(g(x))
(gcircf)(x) = g(f(x))
Let us use the same thing on our problem
f(x) =(-x+1)^(1/2)
g(x) = x^2 - 8
(fcircg)(x)
= (-g(x)+1)^(1/2)
=(-x^2+8+1)^(1/2)
=(-x^2+9)^(1/2)
(fcircg)(x)=(-x^2+9)^(1/2)
Now the other composition
(gcircf)(x)
(gcircf)(x) = (f(x))^2 -8
= ((-x+1)^(1/2))^2-8
= -x+1 - 8
= -x-7
(gcircf)(x) = -x-7
Final answer
(fcircg)(x)=(-x^2+9)^(1/2)
(gcircf)(x) = -x-7
