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

Algebra ->  Functions -> 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      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first two to get you started


1)

a) Domain of f%28x%29=%282%29%2F%28x%5E2%2B1%29

x%5E2%2B1=0 Set the denominator equal to zero. Remember, you cannot divide by zero.

x%5E2=-1 Subtract 1 from both sides


x=sqrt%28-1%29 or x=-sqrt%28-1%29 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 f%28x%29=%282%29%2F%28x%5E2%2B1%29 is all real numbers.


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


b)

Domain of g%28x%29=sqrt%28x-3%29


x-3%3E=0 Set the radicand x-3 greater than or equal to zero


x%3E=3 Add 3 to both sides


So any number greater than or equal to 3 is in the domain of g%28x%29=sqrt%28x-3%29

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




2)

f%28x%29=%282%29%2F%28x%5E2%2B1%29 Start with the first function


f%28g%28x%29%29=%282%29%2F%28%28sqrt%28x-3%29%29%5E2%2B1%29 Plug in g%28x%29=sqrt%28x-3%29. In other words, replace each "x" with sqrt%28x-3%29


f%28g%28x%29%29=%282%29%2F%28%28x-3%29%2B1%29 Square sqrt%28x-3%29 to get x-3. Note: the value of x-3 is now positive. This means that x%3E=3


f%28g%28x%29%29=%282%29%2F%28x-2%29 Combine like terms.


Domain of f(g(x)):

x-2=0 Set the denominator equal to zero. Remember, you cannot divide by zero.

x=2 Add 2 to both sides

So if x=2, then the whole denominator is zero. So this means that we must exclude the value x=2 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