# How to find the range or domain of a function? [closed]

This is a general question I'm asking, I really need it explained. Here's an example of what I mean:

The functions $f$ and $g$ are defined by

$f( x)= x^3 + 1$, $0 ≤ x ≤ 3$

$g(x)= x + 5$, $x \in \mathbb R$.

And I was asked to find the range of $g(f(x))$?? Which I got to be $= x^3 + 6$.

But idk how to find the range and if I was asked to state the domain in another case I wouldn't know how to.

## closed as off-topic by Namaste, Xander Henderson, Jack D'Aurizio, Antonios-Alexandros Robotis, user99914 Oct 8 '17 at 4:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Xander Henderson, Jack D'Aurizio, Antonios-Alexandros Robotis, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

You can find the range of composition of two functions by following the composition step by step. Generally, the range consists of all numbers that the function can produce, given $x$ within a specified interval. First, you find what numbers can be produced by $f$ when given $x$ in $[0,3]$. Then you are to feed the output of $f$ into $g$ and see what $g$ outputs.

It helps to pay attention to the intervals where each function is increasing or decreasing. As a matter of fact, $x^3+1$ is increasing for all $x\ge 0$. Thus, its range on the interval $[0,3]$ is the interval $[f(0),f(3)]$. Which is $[1,28]$.

Next step is to find the range of $g$ on the interval $[1,28]$. Again, it helps that $g$ is increasing. For this reason, the range is $[g(1),g(28)]$. Which is $[6,33]$. This is the range of the composite function as well.

Define $h : \mathbb{R} \rightarrow \mathbb{R}$ as follows: $$h(x) = x^3+6.$$

The important thing is that we've chosen $h$ in such a way that the range of $g \circ f$ just simply equals $h([0,3]).$ So lets go ahead and find $h([0,3])$.

To get started, we need to know the following:

Theorem. Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Then $$f([a,b]) = \left[\mathop{\mathrm{min}}_{x \in [a,b]} f(x), \mathop{\mathrm{max}}_{x \in [a,b]} f(x)\right]$$

Therefore, since $h$ is continuous, we have: $$h([0,3]) = \left[\mathop{\mathrm{min}}_{x \in [0,3]} h(x), \mathop{\mathrm{max}}_{x \in [0,3]} h(x)\right]$$

Now lets use:

Theorem. Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is a function that preserves order on the interval $[a,b]$. Then it has a minimum and a maximum on this interval, and $$\mathop{\mathrm{min}}_{x \in [a,b]} f(x) = f(a), \qquad \mathop{\mathrm{max}}_{x \in [a,b]} f(x) = f(b)$$

Therefore, since $h$ preserves order, we have: $$h([0,3]) = [h(0), h(3)]= [6,33]$$

So the range of $g \circ f$ is $[6,33]$.