SOLUTION: This is a two-part problem. For {{{ f(x) = sqrt ( 1-2x ) }}} {{{ g(x) = -x+3 }}} A) Find (FºG)(x) B) Find the domain of (FºG)(x) What I have so far, which I am not to

Algebra ->  Rational-functions -> SOLUTION: This is a two-part problem. For {{{ f(x) = sqrt ( 1-2x ) }}} {{{ g(x) = -x+3 }}} A) Find (FºG)(x) B) Find the domain of (FºG)(x) What I have so far, which I am not to      Log On


   



Question 287328: This is a two-part problem.
For +f%28x%29+=+sqrt+%28+1-2x+%29+ +g%28x%29+=+-x%2B3+
A) Find (FºG)(x)
B) Find the domain of (FºG)(x)

What I have so far, which I am not totally sure about it:
(FºG)(x) = +sqrt+%28+1-2%28-x%2B3%29+%29+ simplified into +sqrt+%28+2x-5+%29+
Is that correct for step A)? And how do I find the domain for step B)?

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
You did part A exactly right!

For the domain, you just have to find the set of acceptable values for x. The function you found in part A has the x inside a square root. Since we must have a non-negative (zero or positive) number inside the square root, the only values we can allow x to be are those those that make 2x-5 non-negative. In other "words"
2x+-+5+%3E=+0
Solving this:
2x+%3E=+5
x+%3E=+5%2F2
This describes the acceptable values of x. This is the domain.