# Homework Help: Function Definition / Concept - Codomain, range, domain etc.

1. Oct 19, 2012

### binbagsss

What must actually be specified in order for a function to be fully defined / or in what combinations if not all 3 need to be specified?

I.e - from knowing the function you can determine the co-domain - e.g - if it is specified that real functions are going in, and for something simple like 2x + 1 = f(x) - then surely from this you can deduce the co-domain will also be all reals.
- also say if the co-domain and function have both been specified, then given either the domain/range - surely you are able to deduce domain/range accordingly.

- E.g - x^1/2.
Am I correct in thinking that its inverse can only be formed if you either:
- restrict the domain
or
- restrict the codomain

I am pretty certain the 1st option holds, however for the 2nd is my defintion of co-domain correct?

Thanks alot if anyone will shed some light on this, greatly appreciated :).

2. Oct 19, 2012

### Staff: Mentor

For the function f(x) = x1/2, the domain is {x | x≥ 0}. The range is the {y | y ≥ 0}. This function has an inverse, which I choose to write as x = y2. As long as x and y in the latter equation are restricted to exactly the same sets as used in the first equation, then both equations represent one-to-one functions, and both functions have inverses.

Maybe you were thinking of y = x2. This represents a function, but it's not a one-to-one function, so doesn't have an inverse. However, if you restrict the domain to some interval on which the graph is either increasing or decreasing, then on that restricted domain, the function has an inverse.

3. Oct 20, 2012

### 5hassay

Maybe this will help your understanding of a general function $f$:

To define a function, we write

$f : X \rightarrow Y$, $x \in X \mapsto f(x)$,

where $X$ is the domain of $f$, $Y$ is the codomain of $f$. We would define the range of $f$ to be the set $\{f(x) : x \in X\}$.

For example, we wound define $f(x)=x^2$ more formally as the function (considering it only on $ℝ$, the real numbers)

$f : ℝ \rightarrow ℝ^+ \cup \{0\}$, $x \in ℝ \mapsto x^2$

which has domain $ℝ$, codomain $ℝ^+ \cup \{0\}$

Last edited: Oct 20, 2012
4. Oct 24, 2012

### binbagsss

On x^1/2, the domain must be restricted for it to make sense, so not the best example I guess. However once specified ≥ 0, for each x value , there are 2 y values unless you specifically specify the y ≥ 0 . e.g - x=4, y = ±2.

In terms of defining a function is this specification a restriction on the range or co-domain. I believe its the range, and so, so far only the range and domain have been specified. However the codomain has not - is this sufficient to fully define the function, or do we make an assumption on the codomain

Thanks alot anyone.

5. Oct 24, 2012

### Staff: Mentor

No, that's not true. The domain does NOT have to be restricted. The domain is {x | x ≥ 0}. This is not an artificial restriction, because the square root function isn't defined for x < 0.
No again. I assume you're still talking about y = √x = x1/2. For each value of x in the domain, there is exactly one y value.

6. Oct 25, 2012

### binbagsss

Okay thanks I think I understand the idea behind artificial restriction, however my definition of domain must be incorrect - if i take 4 from the domain - are +2 and -2 both not valid y values?

Thanks.

7. Oct 25, 2012

### Staff: Mentor

No. √4 = 2. Period.

Try it on a calculator. Enter 4, then press the √ button. The calculator displays 2. It does not display -2.

Or, look at a graph of y = √x. For each x value in the domain [0, ∞) there is one y value. The point (4, 2) on the graph indicates that √4 = 2. The graph does not have a point (4, -2).

Your confusion is very common. It is true that there are two square roots of 4, the positive (or principal) square root, and a negative square root.

The expression √x is defined to mean the principal square root of x, with x being a nonnegative real number.

(I am referring exclusively here to the function that maps real numbers to real numbers.)