How to find the domain and range of a function with Mathematica?

Question

I'm studying calculus and in some exercises I am asked to find the domain and range of a function. Does Mathematica have already a built-in function for this?

I can imagine some ways of doing so, but at the moment I want to know if there's a built-in function or some easy way of doing it.

Answer 1

Let me just add something about finding the domain:

You can often find the domain of a function f[x] using Reduce. Assuming the domain is a subset of the real numbers, the syntax would be

Reduce[Abs[f[x]] < Infinity, x, Reals]

Here are some examples:

(1)

A piecewise defined function:

Clear[f]; f[x_] := Piecewise[
  {
   {-1.5, x < 0},
   {3, x > 1},
   {Undefined, 0 < x < 1}
   }]

Plot[f[x], {x, -1, 2}]

Reduce[Abs[f[x]] < Infinity, x, Reals]

(* ==> x <= 0 || x >= 1 *)

This means x has to be less than 0 or larger than 1.

(2)

A function with singularities:

Clear[f]; f[x_] := 1/Sin[x]

Plot[f[x], {x, -4 Pi, 4 Pi}]

Reduce[Abs[f[x]] < Infinity, x, Reals]

(*
==> 
C[1] \[Element] 
  Integers && (-Pi + 2 Pi C[1] < x < 2 Pi C[1] || 
   2 Pi C[1] < x < Pi + 2 Pi C[1])
*)

Last[%] /. C[1] -> n

(*
==> -Pi + 2 n Pi < x < 2 n Pi || 
 2 n Pi < x < Pi + 2 n Pi
*)

On the last line I've just made the conditions easier to read by introducing an integer n instead of the default constant C[1].

Answer 2

I think that the closest thing to a built in function, as you ask for, is the the built-in WolframAlpha functionality. For example:

WolframAlpha["domain of x+x/(x(x^2-1))",
 {{"Result", 1}, "Output"}]
WolframAlpha["range of x/(x(x^2-1))",
 {{"Result", 1}, "Output"}]

HoldComplete[x < -1 || -1 < x < 0 || 0 < x < 1 || x > 1]

HoldComplete[y < -1 || y > 0]

Answer 3

Interval is a proper way for your task, but sometimes it is not the way to go. There are functions which can return neither their ranges nor domains, e.g. function yielding k-th non-trivial zero of the Riemman zeta function. E.g. ZetaZero[1000]//N yields 0.5 + 1419.42 I, while N[ ZetaZero @ Interval[{1000, 10000}]] yields ZetaZero[Interval[{1000, 10000}]], however since it is weakly monotonic on the critical line with respect to the imaginary part (i.e. 0.5 + t I, where t is a real number, )

N[ ZetaZero @ {1000, 10000}]

{0.5 + 49.7738 I, 0.5 + 236.524 I}

It is known that ( purely mathematically speaking) the domain of ZetaZero are integer numbers, but there are certain limitations of its implementation in Mathematica, e.g. there is no way to find its domain, only a kind of "divide and conquer" approach, e.g.

ZetaZero[10^7] // N

ZetaZero[10^7 + 1] // N

0.5 + 4.99238*10^6 I

ZetaZero::largp: Argument 10000001 in ZetaZero[10000001] is too large for this 
                   implementation. >>
ZetaZero[10000001]

There is a second argument in ZetaZero[k, t] representing the k-th zero with imaginary part greater than t. Neither this works with Interval, though it is Listable with respect to the second argument :

ZetaZero[105, Interval[{15., 35.5}]]
ZetaZero[105, {15., 35.5}]

ZetaZero[105, Interval[{15., 35.5}]]
{1/2 + 247.137 I, 1/2 + 253.07 I}  

There are similar issues with e.g. PrimePi ( the Mathematica counterpart of the prime counting function $\pi(x)$) and Prime.

PrimePi[ Interval[{10, 100}]]

PrimePi[ Interval[{10, 100}]]   

but

PrimePi[{10, 100}]

{4, 25}

Plot[PrimePi[x], {x, 10, 100}, AxesOrigin -> {0, 0}, PlotStyle -> Thick]

The maximal evaluatable argument for PrimePi is 25 10^13 -1, while with Prime there is a bigger problem, like e.g. an apparently extensible domain :

Prime @ {# + 1, #, # + 1} & 7783516045221

Prime::largp: Argument 7783516045222 in Prime[7783516045222] is too large for
this implementation. >>
{Prime[7783516045222], 249999997909357, 249999997909367}

You can find related details here : What is so special about Prime ?.

There are infinitely many primes, however values of $\pi(x)$ are known up to $x = 10^{23}$, look at e.g. Prime Counting Function on MathWorld .

Answer 4

If you have a domain, you can often find a range using Interval.

Examples:

In[1]:= Sin@Interval[{0, 2 Pi}]
Out[1]= Interval[{-1, 1}]

In[2]:= Sin@Interval[{0, Pi}]   
Out[2]= Interval[{0, 1}]