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The function is $y=2$, the domain is just 2? And the image of it? I don't think I quiet understand what the image of a function means, the domain is all values that it can assume, correct? Could you please try to define the image of this equation too: $y = 2x - 6$, so I can try to understand it? Thank you.

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This really depends on the context of the question. I'm going to make the assumption here that you're discussing functions from the real numbers to the real numbers. Strictly speaking, the domain of the function should be given explicitly when you define the function. However, often one defines a function and then says that the domain is the set of all real numbers for which the function is defined. In this interpretation, the domain of your function $y=f(x)=2$ is all of $\mathbb{R}$ and its image is $\{2\}$. Similarly, the domain of $y=f(x)=2x-6$ is all of $\mathbb{R}$ and its image is $\mathbb{R}$.

To give you a different example, the domain of the function $y=f(x)=\frac{1}{x-1}$ is $\mathbb{R} \setminus \{1\}$ and the image is $\mathbb{R} \setminus \{0\}$.

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The "\" in your answer means "all the real numbers BUT 1", right? Yes, I think I understood it now, thank you very much. So, if the domain of a function is equal to all real numbers then it's image can only be all real numbers as well, correct? –  Luan Cristian Thums May 10 '13 at 12:40
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Yes, the $\setminus$ means subtraction of sets. But your second statement is incorrect. Note that your first example has a domain of $\mathbb{R}$ but an image of $\{2\}$. Another example: the domain of $f(x)=x^2$ is $\mathbb{R}$, but its image is the set of all nonnegative real numbers. –  Alistair Savage May 10 '13 at 12:45
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@LuanCristianThums No! $f(x)=y=2$ for your very own question is a counter-example. It evaluates to 2 for every real number $x$, so it's domain is $\mathbb{R}$, but it's image is $\{2\}$. –  fgp May 10 '13 at 12:45
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@LuanCristianThums Btw, if you had read my answer, you'd have realized that... :-( –  fgp May 10 '13 at 12:46
    
Yes, I understand the logic now, thank you! I was still incorrectly thinking that as it doesn't show a "$x$" it's domain would be just 2 but, yes, just because it "doesn't show" the $X$ I can assume any real number to it and it's image will continue to be $2$. –  Luan Cristian Thums May 10 '13 at 12:48

When you define a function you should always provide a domain (i.e. a start set) and a co-domain (i.e. a target set). Writing $$ f:A\rightarrow B $$ means that $A$ is the domain and $B$ the co-domain. $f$ is a function if it "associates" every single element $a\in A$ with a unique element $b\in B$. We call such $b$ the image of $a$ and denote it by $f(a)$. Then the image of $f$ is the set of all images.

When talking about real valued functions such as $f(x)=2x-6$, it is implied that the domain is the largest subset of $\mathbb R$ for which $f$ is defined (e.g. the domain of $f(x)=\sqrt x$ is $\{x\geq 0\}$), and that the co-domain is $\mathbb R$.

Now, regarding your examples, $y=2$ is short for the (real valued) function $f(x)$ defined as $$ \begin{array}{crcl} f: & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2 \end{array} $$ (note that it is well defined for every $x\in\mathbb R$, so that by the convention above the domain of $f$ is the whole $\mathbb R$). The image of $f$ is just the set $\{2\}$, since no other values of the co-domain are "reached" by $f$.

Concerning $f(x)=2x-6$, since no specification is given, as before we assume $$ \begin{array}{crcl} f: & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2x-6 \end{array} $$ i.e. that the domain is the largest subset of $\mathbb R$ for which the function is defined (so, again the whole $\mathbb R$) and that the co-domain is the whole $\mathbb R$. Now, for example the image of the element 5 is 4, since $f(5)=4$. It is easy to show that the image of $f$ is the whole co-domain $\mathbb R$ (in which case we say that $f$ is surjective, which is not the opposite of injective!). In general, you can find the image of $f$ by "turning $y=f(x)$ in something like $x=\ldots$" where the rhs is an expression that does not contain the variable $x$, and then see for which values of $y$ the equation has sense and the rhs is part of the domain. In this example, $$ y=2x -6 \Rightarrow 2x=y+6 \Rightarrow x= \frac{y+6}2 $$ and the last expression makes sense (i.e. can be computed) for every $y\in\mathbb R$. We then conclude that the image of $f$ is the whole $\mathbb R$.

If we "restricted" the function as follows: $$ \begin{array}{crcl} f: & [0,+\infty) & \longrightarrow & \mathbb R \\ & x & \longmapsto & 2x-6 \end{array} $$ then this time finding its image would result in $$ x= \frac{y+6}2 \quad\text{and}\quad x\geq 0 $$ since now the domain in which $x$ lies is $[0,+\infty)$, i.e. $$ \frac{y+6}2\geq 0 \Leftrightarrow y\geq -6 $$ Therefore, in this case the image of $f$ would be $[-6,+\infty)$.

I hope this example makes it clear.

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Very nice answer. –  Alistair Savage May 10 '13 at 13:32

The image of a function is simply the set of all possible values the function can take. If you have a function $f(x)$, and evaluate it for all possible values of $x$ (i.e., for all $x$ in the domain of f), then the resulting set of values is called the image of $f$.

You can also find a image of a certain subset of the domain - i.e., instead of evalulating $f(x)$ for all $x$ in the domain, you only evaluate for a certain subset.

Formally, the image $f(U)$ of some subset $U$ of the domain of $f$ is $$ f(U) = \left\{f(x) \,:\, x \in U\right\} \text{.} $$

To check whether $y$ lies in the image of $f$, you thus have to ask

Is there an $x$ such that $f(x)=y$?

To check whether $y$ lies in the image of some subset $U$ of the domain of $f$, you similarly ask

Is there an $x \in U$ such that $f(x)=y$?

The image all sets $U$ which aren't empty under $f(x)=2$ is thus simply $\{2\}$, because $f$ never takes any value other than 2.

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An image is a subset of the co-domain with respect to a certain pre-image, which is a subset of the domain. For the function $y=2x-6$, for example, given a pre-image of $[2,10]$, the image is $[-2,14]$. For the function $y=2$, since any input value in the domain will result in $y=2$, besides the null set, the only possible pre-image is ${2}$, and the only possible image is $2$.

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As it was already answered here, if I have none pre-image given then I should assume any real number as a image for $y=2x-6$, right? –  Luan Cristian Thums May 10 '13 at 12:45
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@LuanCristianThums If no pre-image is given, you can assume that the pre-image is the entire domain of the function. And if no domain for the function is given, you can assume it as allistair savage stated. –  Ataraxia May 10 '13 at 12:49

The value of f when applied to x is the image of x under f. y is alternatively known as the output of f for argument x. Now Y=2 is a constant function. The image of Y is always 2 for any value of x. So the Image of function Y=2 is the set containing element 2. Image of 2, under Y=2x−6, is the set containing the element 2(2)-6=4-6=-2.

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