SOLUTION: 1. {{{ f(x)= 2-x^2 }}} {{{ g(x) = sqrt(x+2) }}} I get the following for {{{fg(x) = -x}}}. 2. {{{ g(x) = 1/x }}} I get the following for {{{gg(x) = x}}}. What are the d

Algebra ->  Functions -> SOLUTION: 1. {{{ f(x)= 2-x^2 }}} {{{ g(x) = sqrt(x+2) }}} I get the following for {{{fg(x) = -x}}}. 2. {{{ g(x) = 1/x }}} I get the following for {{{gg(x) = x}}}. What are the d      Log On


   

Question 1167404: 1. +f%28x%29=+2-x%5E2+
+g%28x%29+=+sqrt%28x%2B2%29+
I get the following for fg%28x%29+=+-x.
2. +g%28x%29+=+1%2Fx+
I get the following for gg%28x%29+=+x.
What are the domains?
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Not fg(x). If you are stating function composition then f(g(x)).

#1

You ask about domain.
f%28g%28x%29%29=2-%28sqrt%28x%2B2%29%29%5E2
f%28g%28x%29%29=2-%28x%2B2%29
f%28g%28x%29%29=2-x-2
f%28g%28x%29%29=2-2-x
f%28g%28x%29%29=-x
But look again at the original composition. Necessary is x%3E=-2. Otherwise, f(g(x)) not a "Real" number.