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Can someone please explain to me in layman what each means? Perhaps with some examples with functions (inputs to outputs/numeric values in them)? Especially range, image, and preimage. So far this is my understanding:

  • Domain is basically the input $x$ in $f(x)$.
  • Codomain is what $f(x)$ produces as an output such as $y$ when $f(x) = y$.
  • Range sounds like codomain but with some restriction?
  • Image I have little understanding of but I think it is basically a relation between domain to codomain given that we take a subset of our function (Eg; it is the input to output process of our function given we put a restriction on the domain as $x$ can only go from $0$ to $1$).
  • Preimage is just walking backward on the "image" process? (Inverse image?) Going from our "subsetted" output back to our "subsetted" input?

I am very frustrated that I can't seem to grasp these basic concepts so I would greatly appreciate any help from anyone who can break this down for me and help me understand it without too much mathematical notation. Thank you!

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Consider a function for example $f:R\to R$ defined by $f(x)=x^2$

The domain is the largest possible set of inputs which in this case the set of all real numbers.

The codomain is given as $R$, the set of all real numbers.

The range is the set of all possible outputs which is the interval $[0,\infty)$

The image of a subset $A$ of of real numbers is $f(A)$ which is the set of all $f(x)$ where $x\in A$

For example $f((-1,1))=[0,1)$

The pre-image of a subset $B$ of the range is the set $f^{-1}(B)$ of all imputs $x$ such that $ f(x)$ is in $B$

For example $f^{-1}([1,4])=[-2,-1]\cup [1,2]$

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    $\begingroup$ @Q_A_B_70 Remember that $(-1, 1) = \{x \in \mathbb{R} \mid -1 < x < 1\}$. The square of a number with absolute value less than $1$ must be at least $0$ and less than $1$. Since $f$ is continuous, $f(0) = 0$, and $f(1) = 1$, the function assumes every value between $0$ and $1$, including $0$ but not including $1$ on the interval $[0, 1)$. $\endgroup$ – N. F. Taussig Jun 30 at 13:12
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    $\begingroup$ I think it should be $f(\,(-1,1)\,)$. The outer pair of parentheses being the function application, the inner pair being the open interval. $\endgroup$ – celtschk Jun 30 at 13:29

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