How do you compute (fog) and (gof) if #g(x) = x^2 - 8#, #f(x) = (-x +1)^(1/2)#?

Question

How do you compute (fog) and (gof) if #g(x) = x^2 - 8#, #f(x) = (-x  +1)^(1/2)#?

Elias Ackerman · Accepted Answer

undefined To compute $ (f \circ g) $ and $ (g \circ f) $, first, we'll find the compositions:

1. $ (f \circ g)(x) = f(g(x)) $
2. $ (g \circ f)(x) = g(f(x)) $

Given $ g(x) = x^2 - 8 $ and $ f(x) = (-x + 1)^{\frac{1}{2}} $, we substitute these functions into the compositions:

1. $ (f \circ g)(x) = f(g(x)) = f(x^2 - 8) = \sqrt{-(x^2 - 8) + 1} $
2. $ (g \circ f)(x) = g(f(x)) = g\left((-x + 1)^{\frac{1}{2}}\right) = \left((-x + 1)^{\frac{1}{2}}\right)^2 - 8 = -x + 1 - 8 = -x - 7 $

Cora Alexander · Answer

Step by step working is shown below.

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+1##color(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#