# Math Help - Composite Functions (when you are given an implied domain and range)

1. ## Composite Functions (when you are given an implied domain and range)

Hey guys, I am just confused on these types of questions.

Say if you have a composite function: h(x) = f(g(x)) and you are given an implied domain and an implied range, how do you find out if a number is an element of the domain and range of the outer and inner function?

For example, if the composite function had an implied domain of [2, 6] and implied range of [-1, infinite) how would you find out if 5 is an element of dom(f) and dom(g) and say 6 is an element of ran(g)?

What would be the right method to solve this problem?

2. ## Re: Composite Functions (when you are given an implied domain and range)

Originally Posted by JMoxey
Say if you have a composite function: h(x) = f(g(x)) and you are given an implied domain and an implied range, how do you find out if a number is an element of the domain and range of the outer and inner function?

For example, if the composite function had an implied domain of [2, 6] and implied range of [-1, infinite) how would you find out if 5 is an element of dom(f) and dom(g) and say 6 is an element of ran(g)?
I don't know what "implied domain and an implied range" mean. I have not seen those terms.

However, in order for $f\circ g(a)$ to exist it is necessary that $a\in\text{Dom}(g)~\&~g(a)\in\text{Dom}(f)$.

3. ## Re: Composite Functions (when you are given an implied domain and range)

Originally Posted by Plato
I don't know what "implied domain and an implied range" mean. I have not seen those terms.

However, in order for $f\circ g(a)$ to exist it is necessary that $a\in\text{Dom}(g)~\&~g(a)\in\text{Dom}(f)$.
And also that the range of g(x) is completely contained in the domain of f(x).

4. ## Re: Composite Functions (when you are given an implied domain and range)

Originally Posted by Prove It
And also that the range of g(x) is completely contained in the domain of f(x).
That may not be the case.
Consider $f(x)=\sqrt{x}~\&~g(x)=x^3$.
We can have the function $h(x)=f\circ g(x)=\sqrt{x^3}$ with $\text{Dom}(h)=[0,\infty)=\text{Dom}(f)\cap\text{Rng}(g)$ but $\text{Rng}(g)\not\subset\text{Dom}(f)$.

5. ## Re: Composite Functions (when you are given an implied domain and range)

Ah, no we can't, at least not if we're assuming $\displaystyle f: \mathbf{R} \to \mathbf{R}, f(x) = \sqrt{x}$ and $\displaystyle g: \mathbf{R} \to \mathbf{R}, g(x) = \sqrt{x}$.

$\displaystyle \sqrt{x^3}$ is not defined for any $\displaystyle x < 0$, so that means $\displaystyle f \circ g(x)$ does not exist for any $\displaystyle g(x)$ which are negative. In this case for $f \circ g(x)$ to exist, we require restricting the domain of $\displaystyle g(x)$ so that the range will be $\displaystyle [0, \infty)$.

6. ## Re: Composite Functions (when you are given an implied domain and range)

Originally Posted by Prove It
Ah, no we can't, at least not if we're assuming $\displaystyle f: \mathbf{R} \to \mathbf{R}, f(x) = \sqrt{x}$ and $\displaystyle g: \mathbf{R} \to \mathbf{R}, g(x) = \sqrt{x}$.
$\displaystyle \sqrt{x^3}$ is not defined for any $\displaystyle x < 0$, so that means $\displaystyle f \circ g(x)$ does not exist for any $\displaystyle g(x)$ which are negative. In this case for $f \circ g(x)$ to exist, we require restricting the domain of $\displaystyle g(x)$ so that the range will be $\displaystyle [0, \infty)$.
No one ever said anything about $f:R\to R$. The example simply said $f(x)=\sqrt{x}$ thus its domain is understood to be $[0,\infty)$. We can find this example in almost any Pre-Caculus textbook. It is a standard question to ask about the domain of a composition of two functions.

The domain of $f\circ g$ is $\text{Dom}(f)\cap\text{Rng}(g)$.