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

1 Answer
Jan 14, 2016

Step by step working is shown below.

Explanation:

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#