Let us understand what we need to do for (fcircg)(x)(f∘g)(x) and (gcircf)(x).(g∘f)(x). Before moving ahead let us understand evaluating a function.
Example: Evaluating a function
f(x) = x^2 + 1f(x)=x2+1
Let us evaluate this function at x=2x=2
f(2) = (2)^2+1f(2)=(2)2+1color(red)" Note here " x Note here x color(red)" is replaced by " 2 is replaced by 2
f(2) =4 + 1f(2)=4+1
f(2) = 5f(2)=5
If we have to evaluate this function at x=ax=a then
f(a) = (a)^2+1 f(a)=(a)2+1 color(red)" Note here " x Note here x color(red)" is replaced by " a is replaced by a
If we have to replace it with a function g(x)g(x) then
f(g(x)) = (g(x))^2 + 1f(g(x))=(g(x))2+1
We can see for evaluating we just plug in place of xx
f(g(x))f(g(x)) is same as (fcircg)(x)(f∘g)(x)
(fcircg)(x) = f(g(x))(f∘g)(x)=f(g(x))
(gcircf)(x) = g(f(x))(g∘f)(x)=g(f(x))
Let us use the same thing on our problem
f(x) =(-x+1)^(1/2)f(x)=(−x+1)12
g(x) = x^2 - 8g(x)=x2−8
(fcircg)(x)(f∘g)(x)
= (-g(x)+1)^(1/2)=(−g(x)+1)12
=(-x^2+8+1)^(1/2)=(−x2+8+1)12
=(-x^2+9)^(1/2)=(−x2+9)12
(fcircg)(x)=(-x^2+9)^(1/2)(f∘g)(x)=(−x2+9)12
Now the other composition
(gcircf)(x)(g∘f)(x)
(gcircf)(x) = (f(x))^2 -8(g∘f)(x)=(f(x))2−8
= ((-x+1)^(1/2))^2-8=((−x+1)12)2−8
= -x+1 - 8=−x+1−8
= -x-7=−x−7
(gcircf)(x) = -x-7(g∘f)(x)=−x−7
Final answer
(fcircg)(x)=(-x^2+9)^(1/2)(f∘g)(x)=(−x2+9)12
(gcircf)(x) = -x-7(g∘f)(x)=−x−7
