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The function is

$$g:\Bbb R\setminus\{0\}\to\Bbb R\setminus\{1\}\;,$$

where $$g(x) = x-\frac1x\;.$$

Please pardon my formatting as I am new to this. I know what a function is of course and their domain, codomain and range. What I do not understand is what the Real number part means. Does it mean all real numbers from $0$ to $1$? So it is basically already giving me the domain? Any help would be appreciated.

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In principle the domain and codomain are given explicitly in the first displayed line: the domain ought to be $\Bbb R\setminus\{0\}$, i.e., the set of all non-zero real numbers. Similarly, the codomain is given as $\Bbb R\setminus\{1\}$, the set of all real numbers different from $1$. $\Bbb R$ is the set of all real numbers, positive, negative and $0$. The only part that ought to require any actual work on your part is finding the range: which real numbers different from $1$ are possible values of $g(x)$?

However, on further examination I see that the problem is a bit nasty: there are non-zero real numbers $x$ such that $g(x)=1$, which means that the actual domain of $g$ cannot be all of $\Bbb R\setminus\{0\}$ if $\Bbb R\setminus\{1\}$ is to be the codomain. You can find those numbers by setting $g(x)=1$ and solving for $x$.

Added: This is really a rather bad exercise: properly speaking, the lines

$$g:\Bbb R\setminus\{0\}\to\Bbb R\setminus\{1\}$$

and $$g(x)=x-\frac1x\tag{1}$$

are mutually contradictory, because $\Bbb R\setminus\{1\}$ is not a possible codomain for the function defined by $(1)$ whose domain is $\Bbb R\setminus\{0\}$.

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  • $\begingroup$ So wouldn't that mean that the range is the same as the co-domain in this case? $\endgroup$ – user3554599 Dec 17 '14 at 6:26
  • $\begingroup$ @user3554599: That does turn out to be the case, though you do have to verify it by showing that if $a\ne 1$, then the equation $g(x)=a$ has a solution. $\endgroup$ – Brian M. Scott Dec 17 '14 at 6:28
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The domain is $\mathbb R$ \ $\{0\}$, which mean all real number from $(-\infty, \infty)$ except 0.

The co-domain is $\mathbb R$ \ $\{1\}$, save as above everything except 1.

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A function is mostly defined using the notation $f: A \to B$, which means we can put in objects/elements that 'live' in $A$ and $f$ will produce objects/elements that 'live' in $B$.

The notation $X \setminus Y$ refers to the set of elements that are in $X$ without the elements in $Y$. E.g. $\Bbb{R} \setminus \{2 \}$ means all real numbers except $2$.

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You are partially right. The definition of $g$ will giving you the domain if you rule out all the values of $x$ which satisfies $x-\dfrac{1}{x}=1$. The notation $\mathbb{R}\setminus\{0\}$ simply implies the set of all real numbers excluding $0$.

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