# Thread: Homework Question

1. ## Homework Question

1. Express h(x)=cos(9x^2) as a composition of two functions f and g so that h(x)= f(g(x)).

which I ended up getting
g(x)= 9x^2 and f(x)=cosx

2. Find the domain of the function f(x)= square root of (3x+1)/(5-x)

not positive what to do here, I know that the x cant be 5 since that would make the fraction undefined.

2. Originally Posted by drewms64
1. Express h(x)=cos(9x^2) as a composition of two functions f and g so that h(x)= f(g(x)).

which I ended up getting
g(x)= 9x^2 and f(x)=cosx
That's fine and probably also the solution that you were intended to find.

Originally Posted by drewms64
2. Find the domain of the function f(x)= square root of (3x+1)/(5-x)

not positive what to do here, I know that the x cant be 5 since that would make the fraction undefined.
You have to watch out for two things here:
- denominators cannot become 0 since division by zero isn't defined in R.
- expressions under square roots cannot be negative, squares are nonnegative in R.

I would like to point out though that formally, the domain should be given if you want to talk about a 'function'. Better wording would have been to find the 'maximal' domain on R, sometimes called the 'natural domain'.

3. so for #2 then, the domain would be, x is not equal to 5 and x is greater than or equal to -1/3

one more thing, I got an answer to a problem using a difference quotient where the answer is, 3x^2+3hx+h^2+6 should I factor this further or is that good enoughf or an answer?

thanks for all of the help

4. Originally Posted by drewms64
so for #2 then, the domain would be, x is not equal to 5 and x is greater than or equal to -1/3
Careful! Can x be greater than 5? (I'm assuming the 5-x is also under the square root symbol, yes?)

-Dan

5. yes the whole fraction is under the square root, so the domain would be, x not equal to 5, -1/3<= x< 5

6. Originally Posted by drewms64
yes the whole fraction is under the square root, so the domain would be, x not equal to 5, -1/3<= x< 5
You got it!

-Dan