SOLUTION: Finding (f*g)(x) and (g*f)(x) and the domain of the following. f(X)= squared root of x+1 g(x)= squared root of x-1

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Question 137634This question is from textbook College Algebra Fifth Ed.
: Finding (f*g)(x) and (g*f)(x) and the domain of the following.
f(X)= squared root of x+1
g(x)= squared root of x-1This question is from textbook College Algebra Fifth Ed.

Answer by solver91311(12126) About Me  (Show Source):
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Your terminology is a bit odd, so I'm a little confused as to whether you mean f%28x%29=sqrt%28x%2B1%29 or f%28x%29=%28sqrt%28x%2B1%29%29%5E2. I'll presume the former and you can write back if that isn't correct.

f%28x%29=sqrt%28x%2B1%29

g%28x%29=sqrt%28x-1%29

If you are given a function f, then f(x) means "the value of the function at f"

So if f%28x%29=sqrt%28x%2B1%29, f%280%29=sqrt%280%2B1%29=1, f%283%29=sqrt%283%2B1%29=2, f%28a%29=sqrt%28a%2B1%29. (f°g)(x) is nothing more than f%28g%28x%29%29, so (f°g)(x)= f%28g%28x%29%29=sqrt%28%28sqrt%28x-1%29%29%2B1%29

To find the domain of (f°g)(x), first find the domain of g. The value under the radical cannot be negative, so x-1 must be non-negative, in other words x-1%3E=0, or x%3E=1. Which is to say that anything less than 1 must be excluded.

Now let's examine the domain of f. Using similar analysis, the domain of f is x%3E=-1

So for the composite function, we have to restrict g(x) to be greater than -1.

In other words, g%28x%29=sqrt%28x-1%29%3E=-1.

Since g(x) is positive for all values of x in the domain of g, g(x) is also greater than or equal to -1 for all values of x in the domain of g.

Hence, the only restriction on (f°g)(x) is that x%3E=1, and therefore the domain of (f°g)(x) is the interval [1,infinity)

You should be able to handle the other part of this problem now.