Expert: Steve Holleran - 7/22/2008

Question
a. Write the formula for f(g(x))and g(f(x))and find the
b. domain and
c. range of each.
1. f(x) = √x
  g(x) = √(1 - x)

2. f(x) = 2 - x^2
  g(x) = √(x + 2)

Answer
Hi Davis,

These compositions mean that you want to substitute each function into the other .   Always work from the inside out, so

f(g(x)) means to put the g function into the f function.  Since f takes the square root of whatever you are inputting, f(g(x)) will take the square root of rt(1-x), which means you have

          f(g(x)) = rt[rt(1-x)] = (1-x)^1/4

For the domain, you have to do g first, so 1-x>=0 , and 1>=x or x<=1
Since the outputs here are all >=0, the range is y>=0.

For g(f(x)), put rt(x) into g, so you have

          g(f(x)) = rt[1-rt(x)]

Now, for the domain, you can put any x >=0 into f, but the outputs are also >=0.  The problem is when the f outputs become >1, then

                      rt[1 - rt(x)] tries to take a root of a negative number.  So, only x's between 0 and 1 can be put into g(f).


For the second one, f(g) = f(rt(x+2)) = 2 - [rt(x+2)]^2 = 2 - (x+2)

                                     = -x

but this has to have domain x>=-2, because any other x's will make negative radicands in g.  The range then is y>=2.

Then g(f) = g(2-x^2) = rt[2-x^2].  Now f has a domain of all reals, but the outputs are all <=2.  These now become the inputs into g, but g won't take the root of a negative result, so 2-x^2 must be >=0,
which means x>=-2 and x<=2 is the domain.  The range would be positive results, but only from o to rt(2), since the largest output of f is 2, then the largest output of g(f) has to be rt(2)

Hope this helps
Steve