## Simple function questions - checking answer.

1. The problem statement, all variables and given/known data

1) Given f(x)=x+3, x∈R and g(x)=x2,x∈R,0=<x=<4, find the range of composite function fg(x).

2) Given f(x)=x+1,x∈R,x>=5 and g(x)=x1/2, x∈R,x>=0. Find the domain of fg(x).

2. Relevant equations

3. The attempt at a solution

For 1), the answer 1 get is y=>3 , but the answer given is3=<y=<19.

2) the answer i get is x>=0 , but yet the answer provided is x>=25.

So what answers do you guys get? Anything wrong with the answers provided by the answer sheet? Cause i couldn't find anything wrong with my solutions... Thanks in advance.
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 You almost had #1. You just didn't handle the other part of the inequality! If you write out your work we can find where you went wrong.

Recognitions:
Homework Help
 Quote by Michael_Light 2) Given f(x)=x+1,x∈R,x>=5 and g(x)=x1/2, x∈R,x>=0. Find the domain of fg(x). 2) the answer i get is x>=0 , but yet the answer provided is x>=25.
The domain of f(g(x)) consists of all x-values is such that the g(x)-values maps to the domain of f(x). It can't be x ≥ 0 because the range of g(x) would be g(x) ≥ 0, and you would have values that are not permitted in the domain of f(x).

If you restrict the domain of f(g(x)) to be x ≥ 25, then the range of g(x) would be g(x) ≥ 5, which "matches" the domain of f(x). So x ≥ 25 is right.

## Simple function questions - checking answer.

 Quote by QuarkCharmer You almost had #1. You just didn't handle the other part of the inequality! If you write out your work we can find where you went wrong.
I finally managed to get the correct answer for 1), can you help me with 2)? ><

 Quote by eumyang The domain of f(g(x)) consists of all x-values is such that the g(x)-values maps to the domain of f(x). It can't be x ≥ 0 because the range of g(x) would be g(x) ≥ 0, and you would have values that are not permitted in the domain of f(x). If you restrict the domain of f(g(x)) to be x ≥ 25, then the range of g(x) would be g(x) ≥ 5, which "matches" the domain of f(x). So x ≥ 25 is right.
But how can i obtain x≥ 25 mathematically?Here is my approach... fg(x) = f(x1/2) = x1/2+1, x≥0. Domain of fg(x)= domain of g(x), hence domain of fg(x) is x≥0. Can anyone correct me?

Mentor
 Quote by Michael_Light I finally managed to get the correct answer for 1), can you help me with 2)? >< But how can i obtain x≥ 25 mathematically?Here is my approach... fg(x) = f(x1/2) = x1/2+1, x≥0. Domain of fg(x)= domain of g(x), hence domain of fg(x) is x≥0. Can anyone correct me?
You really should write this as f(g(x)), not fg(x), which is what is used for the product of two functions.

You are given that the domain of f is restricted to x >= 5. So although there are many values of x that are in the domain of g, there are some values of g(x) that aren't in the domain of f.

Some examples that use selected values of x.
x = 0. g(0) = sqrt(0) = 0. Can we evaluate f(0)? No, because 0 is not >= 5.
x = 1. g(1) = sqrt(1) = 1. Can we evaluate f(1)? No, because 1 is not >= 5.
x = 4. g(4) = sqrt(4) = 2. Can we evaluate f(2)? No, because 2 is not >= 5.
x = 9. g(9) = sqrt(9) = 3. Can we evaluate f(3)? No, because 3 is not >= 5.
Do you see where I'm going with this?

Mentor
 Quote by Michael_Light ... But how can i obtain x≥ 25 mathematically?Here is my approach... fg(x) = f(x1/2) = x1/2+1, x≥0. Domain of fg(x)= domain of g(x), hence domain of fg(x) is x≥0. Can anyone correct me?
For #2:

Since f(x) is defined only for x ≥ 5, find the values of x for which g(x) ≥ 5.