SOLUTION: I am having touble with this F of G and G of F. A) Determine the domain and range of a) F(x)= 2/(x^2+1) and b) g(x)=Square root of x-3 B) Find the formula for f(g(x)) and the d

Question 155654: I am having touble with this F of G and G of F.
A) Determine the domain and range of a) F(x)= 2/(x^2+1) and b) g(x)=Square root of x-3
B) Find the formula for f(g(x)) and the domain of f(g(x))
C) Find the formula for g(f(x)) and the domain of g(f(x)).
Thankyou so much...
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you started

1)

a) Domain of

Set the denominator equal to zero. Remember, you cannot divide by zero.

Subtract 1 from both sides

or Take the square root of both sides

Since the square root of negative 1 is not a real number, this means that no real number will make the denominator equal to zero. So there are no domain restrictions. This means that the domain of is all real numbers.

So the domain of f(x) in set-builder notation is

which is (

) in interval notation

b)

Domain of

Set the radicand greater than or equal to zero

Add 3 to both sides

So any number greater than or equal to 3 is in the domain of

So the domain of g(x) in set-builder notation is

which is [

) in interval notation

2)

Start with the first function

Plug in . In other words, replace each "x" with

Square to get . Note: the value of is now positive. This means that

Combine like terms.

Domain of f(g(x)):

Set the denominator equal to zero. Remember, you cannot divide by zero.

Add 2 to both sides

So if , then the whole denominator is zero. So this means that we must exclude the value from the domain. However, we specified earlier that x must be greater than or equal to 3. Since 2 is not greater than or equal to 3, we don't have to worry about restricting this value (as it has been already done)

So the domain of f(g(x)) in set-builder notation is

which is [

) in interval notation

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