Function Definition / Concept - Codomain, range, domain etc.

In summary: 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.
  • #1
binbagsss
1,259
11
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 a lot if anyone will shed some light on this, greatly appreciated :).
 
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  • #2
binbagsss said:
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
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.


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

Thanks a lot if anyone will shed some light on this, greatly appreciated :).
 
  • #3
Maybe this will help your understanding of a general function [itex]f[/itex]:

To define a function, we write

[itex]f : X \rightarrow Y[/itex], [itex]x \in X \mapsto f(x)[/itex],

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

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

[itex]f : ℝ \rightarrow ℝ^+ \cup \{0\}[/itex], [itex]x \in ℝ \mapsto x^2[/itex]

which has domain [itex]ℝ[/itex], codomain [itex]ℝ^+ \cup \{0\}[/itex]
 
Last edited:
  • #4
Mark44 said:
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.

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 a lot anyone.
 
  • #5
binbagsss said:
On x^1/2, the domain must be restricted for it to make sense, so not the best example I guess.
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.
binbagsss said:
However once specified ≥ 0, for each x value , there are 2 y values
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.
binbagsss said:
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
 
  • #6
Mark44 said:
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.

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
binbagsss said:
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?
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.)
 

Related to Function Definition / Concept - Codomain, range, domain etc.

1. What is a function?

A function is a mathematical rule or relationship between two sets of numbers, where each input (or independent variable) has exactly one output (or dependent variable). It can also be thought of as a machine that takes in an input and produces an output.

2. What is a domain?

The domain of a function is the set of all possible input values for that function. It is the independent variable in the function and determines what values can be used as inputs.

3. What is a codomain?

The codomain of a function is the set of all possible output values for that function. It is the dependent variable in the function and determines what values can be produced as outputs.

4. What is the difference between range and codomain?

The range of a function is the set of all actual output values that are produced when the function is evaluated with different inputs. It is a subset of the codomain and may be smaller than the codomain if not all values in the codomain are actually produced by the function.

5. Can a function have multiple outputs for a single input?

No, by definition, a function can only have one output for each input. This is known as the vertical line test - if a vertical line can intersect the graph of a function more than once, then it is not a function.

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