Question - how to find the domain of a composite function.
Thanks for the A2A. Let’s consider a composition h(x) = g(f(x)):
x —→ f(x) —→ g(x)
First, the domain of h is a subset of the domain of f, because in order to get h(x) = g(f(x)), you first have to be able to have f(x). Second, the number f(x) must be in the domain of g. In other words, the range of f is a subset of the domain of g. Implicit in that, however, is we need to look at the domain of g as well.
Putting those together, we appear, at first look, to get the result that
dom(h) = dom(f) intersect dom(g). Is that sufficient? Let’s see.
E
Question - how to find the domain of a composite function.
Thanks for the A2A. Let’s consider a composition h(x) = g(f(x)):
x —→ f(x) —→ g(x)
First, the domain of h is a subset of the domain of f, because in order to get h(x) = g(f(x)), you first have to be able to have f(x). Second, the number f(x) must be in the domain of g. In other words, the range of f is a subset of the domain of g. Implicit in that, however, is we need to look at the domain of g as well.
Putting those together, we appear, at first look, to get the result that
dom(h) = dom(f) intersect dom(g). Is that sufficient? Let’s see.
Example 1: f(x) = 1/x, g(x) = sqrt(x), so that h(x) = sqrt(1/x). The domain of f is the set of all real numbers except 0. Hence, x = 0 is the only value excluded so far. The domain of g is the set of all real numbers that are >= 0. The intersection of the two sets (domain(f), domain(g)) is the set of positive real numbers. OK.
Example 2: f(x) = 1/x, g(x) = 1/x, so that h(x) = 1/(1/x) = x, but equaling x is only for x nonzero. On the surface, it appears that the domain of h could be the set of all real numbers, but x = 0 could never be “put into” the first function f. The domain of h is the set of all nonzero real numbers. Still good.
Example 3: f(x) = 1/x, g(x) = 1/(x-1), so that h(x) = 1 / (f(x) - 1) = 1 / (1/x - 1) . To simplify that further, multiply top and bottom by x to get x / (1 - x), which, at first blush, appears to avoid only x = 1. However, as in the previous example, x also cannot equal 0. So, the domain of h is the set of all reals except 0, 1. OK.
Example 4: Here is where the obvious conjecture above goes south!
Let f(x) = 4, g(x) = 1/x, so that h(x) = 1/4 for all real numbers x. Note that x = 0 is NOT in the domain of g, but since f(x) is never 0, it’s NOT a problem. It is irrelevant that the domain of g has a restriction! So, in this case, we see that the domain of h is the set of ALL real numbers.
The moral of the last example? It shows how the range of f must be taken into account. Yes, we need to include only those numbers that are in the domain of f, but only those for which the elements of the range of f are in the domain of g. Let us write g* for the restriction of function g to only those values also in the range of f. To be clear, g* would be the same as g, BUT ONLY FOR NUMBERS that are in the range of f. Our refined conclusion is that the domain of h is:
the intersection of dom(f) and dom(g*).
The range and domain of composite functions.
Consider the composite function [math]g(f(x)).[/math]
What do we do here? We first take the variable [math]x[/math] and subject it to the function [math]f.[/math] We then take the output of the function [math]f[/math] and subject it to the function [math]g[/math] to get the final output.
[math]\Rightarrow\qquad[/math] The domain of [math]g(f(x))[/math] is the domain of [math]f[/math] since that is the function which operates on the variable [math]x[/math] and the range of [math]g(f(x))[/math] is the range of [math]g[/math] since that is the range of the output.
One point to be noted is that the range of the first function, [math]f,[/math] should be a subset of the domain of the second function, [math]g[/math] since tha
The range and domain of composite functions.
Consider the composite function [math]g(f(x)).[/math]
What do we do here? We first take the variable [math]x[/math] and subject it to the function [math]f.[/math] We then take the output of the function [math]f[/math] and subject it to the function [math]g[/math] to get the final output.
[math]\Rightarrow\qquad[/math] The domain of [math]g(f(x))[/math] is the domain of [math]f[/math] since that is the function which operates on the variable [math]x[/math] and the range of [math]g(f(x))[/math] is the range of [math]g[/math] since that is the range of the output.
One point to be noted is that the range of the first function, [math]f,[/math] should be a subset of the domain of the second function, [math]g[/math] since that is the input for [math]g.[/math]
Continuous functions and open mappings are very different things, although the definitions seem kind of similar. It’s really important to understand the significance and nature of these kinds of functions, beyond the dry definitions.
If [math]f:X\to Y[/math] is a function between two topological spaces, we say that
- [math]f[/math] is continuous if [math]f^{-1}(U)[/math] is an open set (of [math]X[/math]) whenever [math]U[/math] is an open set (of [math]Y[/math])
- [math]f[/math] is open if [math]f(U)[/math] is an open set (of [math]Y[/math]) whenever [math]U[/math] is an open set (of [math]X[/math]).
The notation [math]f(U)[/math] simply means the set of all points [math]f(x)[/math] where [math]x\in U[/math]. Similarly, the set [math]f^{-1}(U)[/math] is the set of all points [math]x[/math] such that [math]f(x)[/math]
Continuous functions and open mappings are very different things, although the definitions seem kind of similar. It’s really important to understand the significance and nature of these kinds of functions, beyond the dry definitions.
If [math]f:X\to Y[/math] is a function between two topological spaces, we say that
- [math]f[/math] is continuous if [math]f^{-1}(U)[/math] is an open set (of [math]X[/math]) whenever [math]U[/math] is an open set (of [math]Y[/math])
- [math]f[/math] is open if [math]f(U)[/math] is an open set (of [math]Y[/math]) whenever [math]U[/math] is an open set (of [math]X[/math]).
The notation [math]f(U)[/math] simply means the set of all points [math]f(x)[/math] where [math]x\in U[/math]. Similarly, the set [math]f^{-1}(U)[/math] is the set of all points [math]x[/math] such that [math]f(x)\in U[/math].
Those are the formal definitions, and they’re clearly different, but what do they mean?
A continuous function takes nearby points to nearby points. It’s not allowed to tear thing apart, sending point [math]a[/math] to point [math]b[/math] and things arbitrarily close to [math]a[/math] somewhere away from [math]b[/math].
On the other hand, a continuous map is allowed to smash things together. Nothing wrong with that. In fact, one of the simplest of continuous functions is a constant function such as [math]f(x)=7[/math] as a function from [math]\R[/math] to [math]\R[/math].
In fact, for a constant map, the inverse image [math]f^{-1}(A)[/math] for any set [math]A[/math] is either empty or the whole space, depending on whether or not [math]A[/math] contains the constant value. Both the empty set and the whole space are open sets, so continuity is guaranteed.
Open maps, on the other hand, are not allowed to smash stuff together. They “retain breathing space”, moving each open set to a set that’s still open. A single point is not an open set in [math]\R[/math] or [math]\R^n[/math] or most other topological spaces, so a constant map – which we just saw was continuous – is most definitely not open.
Let’s look at a more interesting example of a continuous map that’s not open: [math]f(x)=x^2[/math], once again as a function from [math]\R[/math] to [math]\R[/math].
It should be intuitively clear that [math]f[/math] is continuous: nearby numbers have nearby squares. Let’s see this from the perspective of open sets and inverse images:
- The inverse image of [math](-2,-1)[/math] is empty. No real number has a negative square. Empty sets are open. Check.
- The inverse image of [math](-4,4)[/math] is [math](-2,2)[/math]. Open interval. Check.
- The inverse image of [math](4,9)[/math] is [math](-3,-2)\cup (2,3)[/math], the union of two disjoint open intervals. That’s an open set too, so check.
Those examples are typical. The set of numbers mapped by [math]f[/math] into an open interval is either empty, or an interval, or two intervals. All of these are open sets, so the topological definition of continuity is satisfied.
On the other hand, [math]f[/math] is not an open map. The forward image [math]f(I)[/math] of the interval [math]I=(-1,1)[/math] is the half-open interval [math][0,1)[/math], which is not an open set. [math]f[/math] “folds” that interval on itself, creating a sharp edge at [math]0[/math], and that breaks openness.
We’ve seen some examples of continuous maps that aren’t open: constant maps, and the map [math]x\mapsto x^2[/math] which failed to be open around [math]0[/math]. Are the open maps that aren’t continuous?
There certainly are, but constructing them as functions between the “usual” Euclidean spaces is somewhat tricky. One example is the Conway base 13 function, which receives every real value in any interval, however small. It is wildly discontinuous, but certainly open, because the image of any open set is the entire real line.
We can construct a somewhat less wild example, but we’ll need to change the range a bit artificially. Consider the function
[math]\displaystyle f:\R \to [-1,1][/math]
[math]\displaystyle f(x)=\begin{cases}\sin(1/x) & \text{if $x \neq 0$} \\ 0 & \text{if $x=0$}\end{cases}[/math]
Notice that the codomain of [math]f[/math] is not [math]\R[/math] but merely the closed interval [math][-1,1][/math]. Within that space, an interval such as [math](a,1][/math] is actually an open set, and of course the entire space [math][-1,1][/math] is open as well. As a result, the forward image [math]f(I)[/math] of any open interval is open.
On the other hand, [math]f[/math] is of course not continuous. The inverse image [math]f^{-1}(I)[/math] of any open interval [math]I[/math] containing [math]0[/math], like for example [math]I=(-\frac{1}{10},\frac{1}{10})[/math], is the union of the single point [math]\{0\}[/math] and infinitely many tiny open intervals – which is not an open set. The function is not continuous.
Continuous maps are the key actors in the entire field of topology. The category of topological spaces is exactly made up of topological spaces and continuous maps between them.
Open maps are certainly not as ubiquitous as continuous maps, but they are important in numerous contexts. For example, a fundamentally important topological property of holomorphic maps is that they are either constant or open (we just saw that this isn’t true for smooth real maps).
Another crucial example is the behavior of injective (one-to-one) continuous maps [math]f:\R^n\to\R^n[/math]. A fundamental result known as invariance of domain shows that such maps are open. Contrast this with the example [math]x \to x^2[/math] we studied, which failed to be open precisely because it failed to be injective. It seems almost obvious that injective maps must retain the “volume” of sets they transform, so they should be open. Surprisingly, invariance of domain is a quite difficult theorem to prove, requiring tools from algebraic topology.
If a function F is f(g(x)), I learned that domain restrictions of g(x) apply to domain of F, why is that? Why does a function being a composite change its domain?
In [math]f(g(x))[/math], [math]f[/math] must be able to accept any value the [math]g[/math] outputs. There is no extra restriction on [math]x[/math], but if the domain of [math]f[/math] is a proper subset of the range of [math]g[/math], the composite function is not defined.
For example if [math]g(x)=x^3[/math] with domain the real numbers, and [math]f(x)=\sqrt{x}[/math], with domain the positive real numbers, then [math]f(g(x))=\sqrt{x^3}[/math] is undefined because it doesn’t know what to do with a negative [math]x[/math] value.
P.S. it doesn’t matter if the dom
If a function F is f(g(x)), I learned that domain restrictions of g(x) apply to domain of F, why is that? Why does a function being a composite change its domain?
In [math]f(g(x))[/math], [math]f[/math] must be able to accept any value the [math]g[/math] outputs. There is no extra restriction on [math]x[/math], but if the domain of [math]f[/math] is a proper subset of the range of [math]g[/math], the composite function is not defined.
For example if [math]g(x)=x^3[/math] with domain the real numbers, and [math]f(x)=\sqrt{x}[/math], with domain the positive real numbers, then [math]f(g(x))=\sqrt{x^3}[/math] is undefined because it doesn’t know what to do with a negative [math]x[/math] value.
P.S. it doesn’t matter if the domain of [math]f[/math] is a proper superset of the range of [math]g[/math].
The parameter method
If you have each function given explicitly, say f(x) = x+3 and g(x)=x^2 and you want to find the domain and range of g(f(x)) then the easiest thing to do is form a function in one variable by passing the parameter through. So g(f(x))=g(x+3)=(x+3)^2. Finding the domain and range of this is simple, it's just a normal function. This method applies for higher orders of composition too. However, if you don't have the functions explicitly (we’re not always that lucky) or want to think about it in terms of sets read below.
The set method
Let's say our composite function is g(f(x)).
The parameter method
If you have each function given explicitly, say f(x) = x+3 and g(x)=x^2 and you want to find the domain and range of g(f(x)) then the easiest thing to do is form a function in one variable by passing the parameter through. So g(f(x))=g(x+3)=(x+3)^2. Finding the domain and range of this is simple, it's just a normal function. This method applies for higher orders of composition too. However, if you don't have the functions explicitly (we’re not always that lucky) or want to think about it in terms of sets read below.
The set method
Let's say our composite function is g(f(x)). The easiest way to determine domain and range is to split the functions up.
Start by taking f(x). Determine the range of that function. Take g(x) and determine it's domain. Do this as if they were individual functions - this should be fairly easy.
Now take the intersection of those two sets, call it X. This is the set that links the two functions, it is mapable from the domain of g(f(x)) by carrying out f, and its mapable to the range of g(f(x)) by carrying out g.
The domain of g(f(x)) is the preimage of X in f, or if you prefer f^-1(X)=Y. The range of g(f(x)) is g(X)=Z.
This forms the sequence of equivalences g(f(Y))=g(X)=Z. So Y is the domain and Z is the range. This should be clear by f: Y → X and g: X → Z.
This process can be extended to more highly composite functions such as f(g(h(x))) but requires a bit more moving backwards and forwards. I would start by determining the domain and range of g(h(x)) as above. Then carry out the process of intersecting the domain of f with the range of g(h(x)), call it X. You should then be fairly easily be able to get the total range by calculating f(X)=W. You'll then need to work back to determine the final domain of g(h(x)) that is allowable. To do this you carry out the preimage of g on X, I.e. g^-1(X) = Y. Then carry out the preimage of h on Y, I.e h^-1(Y) = Z. This gives the final domain.
Written more nicely we will have found h: Z → Y, g: Y → X and f: X → W where Z is the domain and W is the range of f(g(h(x))).
I hope this explanation helps. The set methodology requires a bit of thought but after some examples it becomes easier. If anyone has any examples they need help with feel free to comment or request me to answer another question and I'll get round to it.
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Consider a function [math]f:[/math][math] \, A \, \rightarrow \, B[/math] and another function [math]g:[/math][math] \, B \, \rightarrow \, C[/math].
Then, we have a composite function [math]g \circ f: \, A \, \rightarrow \, C[/math]
So, [math]y \, =\, f(x) \qquad z \, =\, g(y) \qquad and \qquad z \, =\, g \circ f(x) [/math]
The domain of the composite function is the domain of f (i.e set A) and the range of the composite function is the range of g (i.e set C).
A function [math]f \colon X \to Y[/math] is continuous if and only if the inverse image (or pre-image) of each open set is open, ie, for each open set [math]V \subseteq Y[/math], its pre-image [math]f^{-1}(V)[/math] is open in [math]X[/math].
A function [math]f \colon X \to Y[/math] is open if and only if the direct image of each open set is open, ie
for each open set [math]U \subseteq X[/math], its image [math]f(U)[/math] is open in [math]Y[/math].
The question does not seem to impose any restrictions on the topologies.
This way, it is easy to construct very simple applications that are open but not continuous.
Just take the identity function, and let the spaces [math]X[/math] and [math]Y[/math] have the same base set but th
A function [math]f \colon X \to Y[/math] is continuous if and only if the inverse image (or pre-image) of each open set is open, ie, for each open set [math]V \subseteq Y[/math], its pre-image [math]f^{-1}(V)[/math] is open in [math]X[/math].
A function [math]f \colon X \to Y[/math] is open if and only if the direct image of each open set is open, ie
for each open set [math]U \subseteq X[/math], its image [math]f(U)[/math] is open in [math]Y[/math].
The question does not seem to impose any restrictions on the topologies.
This way, it is easy to construct very simple applications that are open but not continuous.
Just take the identity function, and let the spaces [math]X[/math] and [math]Y[/math] have the same base set but the topology in [math]X[/math] be coarser than that in [math]Y[/math].
For example let [math]X[/math] be the reals with the ordinary topology, and [math]Y[/math] be the reals with the discrete topology (in which all sets are open). Then obviously the identity map is open (since the image of any set is open), but it is not continuous (the pre-image of the set [math]\{0\}[/math], which is open in the discrete topology, is again [math]\{0\}[/math], and is not open in the ordinary topology).
It is a bit more difficult to construct examples if one requires to topological spaces on either side to be the same.
One such example, where the topology on both sides is the usual topology on the reals can be seen here:
This is basically a function of a function or a mapping of a mapping.
In this diagram, [math]g[/math] maps points from the domain of [math]g[/math] to the range of [math]g[/math]. Then f maps those same points from the range of [math]g[/math] (which is part of the domain of [math]f[/math]) to the range of [math]f[/math].
In functional notation, we write [math](fog)(x)[/math] or [math]f(g(x))[/math] or [math]f(u)[/math] where [math]u = g(x)[/math].
That is, whenever you see an [math]x[/math] in the equation [math]f(x)[/math] you replace it with [math]g(x)[/math].
For example, if [math]f(x) = \sqrt{x+5} - 2x[/math] and [math]g(x) = 2-x^2[/math] then
[math](fog)(x) = \sqrt{2-x^2 +5} - 2(2-x^2) = \sqrt{7-x^2} -4x^2–10[/math]
Note that for the function to be described by this composite mapping, the range of [math][/math]
This is basically a function of a function or a mapping of a mapping.
In this diagram, [math]g[/math] maps points from the domain of [math]g[/math] to the range of [math]g[/math]. Then f maps those same points from the range of [math]g[/math] (which is part of the domain of [math]f[/math]) to the range of [math]f[/math].
In functional notation, we write [math](fog)(x)[/math] or [math]f(g(x))[/math] or [math]f(u)[/math] where [math]u = g(x)[/math].
That is, whenever you see an [math]x[/math] in the equation [math]f(x)[/math] you replace it with [math]g(x)[/math].
For example, if [math]f(x) = \sqrt{x+5} - 2x[/math] and [math]g(x) = 2-x^2[/math] then
[math](fog)(x) = \sqrt{2-x^2 +5} - 2(2-x^2) = \sqrt{7-x^2} -4x^2–10[/math]
Note that for the function to be described by this composite mapping, the range of [math]g(x)[/math] must be a subset of the domain of [math]f(x)[/math]. This may mean restricting the domain of [math]g(x)[/math]. In this example, there is no restriction on the domain of [math]g(x)[/math] but for the composite function, the domain of [math]g(x)[/math] must be restricted to [math]-\sqrt{7} \le x \le \sqrt{7}[/math]
The domain of [math]g(x)[/math] is the largest possible domain of [math]fog(x)[/math] even if there is no apparent restriction otherwise when [math]fog(x)[/math] is simplified. For example, if [math]g(x) = \sqrt{x}[/math] and [math]f(x)= x^2+3[/math] then [math](fog)(x) = x+3[/math]. Although it appears that there is no restriction on [math]fog(x)[/math], the domain of [math]g(x)[/math] is [math]x \ge 0[/math] so the domain of [math](fog)(x)[/math] must also be [math]x \ge 0[/math] otherwise [math]x+3[/math] can’t be described by this composite mapping.
A common function usually has as domain a subset of the real numbers. You can pick any numbers from the domain to replace a variable.
A Composite function has its domain determined by another function not a set of numbers. we call it " a function of a function. Example :
If : . f (x ) = 3x + 7 . and . g (x) = 5x - 5. . find and simplify f [ g(x ) ] this is a composite function because the domain is a function of another function., f (x ) is a simple function or just a function .
Lets us simplify the example I wrote above.
f [ g (x ) means " Go to the function f (x ) and replace x by g (x). . f [ g
A common function usually has as domain a subset of the real numbers. You can pick any numbers from the domain to replace a variable.
A Composite function has its domain determined by another function not a set of numbers. we call it " a function of a function. Example :
If : . f (x ) = 3x + 7 . and . g (x) = 5x - 5. . find and simplify f [ g(x ) ] this is a composite function because the domain is a function of another function., f (x ) is a simple function or just a function .
Lets us simplify the example I wrote above.
f [ g (x ) means " Go to the function f (x ) and replace x by g (x). . f [ g (x) ] is a Composite function
we have : f [ g (x) ] = 3 g(x) + 7 . but . g( x) = 5x - 5
replace x in f[ g(x ) ] . by . 5x - 5 and simplify
f [ g(x) ] = 3 ( 5x - 5) + 7 = 15x - 8 **
In applied math, like in electronics , g(x) is called " a filter " , because it filters the numbers that will become the domain of f (x). For example:
The function f (x) = 2/ ( x + 2 ) ,will became undefined if x is replaces by negative -2 because the denominator becomes zero. But if we have the composite function :
f [ g (x) ] = 2/ ( x + 2) . and . g ( x ) = | x | then
f [ g (x ) ] = 2 / ( | x | + 2 ) and if x is replaced by -2 ,we have no problem.
Now a common event.
In an elegant restaurant the waiter gives to his busboy 10 % of how much the waiter gets in tips in one day .That depends on how much the customer give him. ( that is a simple function. The domain is made by the set of customers ). How about, how much the busboy get ? That depends on how much the waiter gets in tips, which depends on how much tips the customer gives the waiter ( That is a Composite function )
Let [math]y = f(u)[/math] and [math]u = g(x)[/math]
[math]\Rightarrow \qquad y = f(g(x))[/math]
Then, the domain of the function [math]y = f(g(x))[/math] is the domain of the function [math]u = g(x)[/math] and the range of the function [math]y = f(g(x))[/math] is the range of the function [math]y = f(u).[/math]
In this particular case, [math]f(x) = 2x + 1, x > 1[/math] and [math]g(x) = x^2, x > 3.[/math]
The domain and range of [math]f(x)[/math] are [math](1, \infty)[/math] and [math](3, \infty)[/math] respectively.
The domain and range of g[math](x)[/math] are [math](3, \infty)[/math] and [math](9, \infty)[/math] respectively.
The domain of f[math](g(x))[/math] is the domain of [math]g(x)[/math], which is [math](3, \infty)[/math]. The range of [math] g(x)[/math], which is [math](9, \infty)[/math] is then to be considered as the restricted domain of [math]f(x)[/math]
Let [math]y = f(u)[/math] and [math]u = g(x)[/math]
[math]\Rightarrow \qquad y = f(g(x))[/math]
Then, the domain of the function [math]y = f(g(x))[/math] is the domain of the function [math]u = g(x)[/math] and the range of the function [math]y = f(g(x))[/math] is the range of the function [math]y = f(u).[/math]
In this particular case, [math]f(x) = 2x + 1, x > 1[/math] and [math]g(x) = x^2, x > 3.[/math]
The domain and range of [math]f(x)[/math] are [math](1, \infty)[/math] and [math](3, \infty)[/math] respectively.
The domain and range of g[math](x)[/math] are [math](3, \infty)[/math] and [math](9, \infty)[/math] respectively.
The domain of f[math](g(x))[/math] is the domain of [math]g(x)[/math], which is [math](3, \infty)[/math]. The range of [math] g(x)[/math], which is [math](9, \infty)[/math] is then to be considered as the restricted domain of [math]f(x)[/math] for considering the range of [math]f(g(x))[/math]. With this restricted domain, the range of [math]f(x)[/math] is restricted to [math](19, \infty).[/math]
So, the domain and range of [math]f(g(x))[/math] are [math](3, \infty)[/math] and [math](19, \infty)[/math] respectively.
The domain of [math]g(f(x))[/math] is the domain of [math]f(x)[/math], which is [math](1, \infty).[/math] The range of [math]f(x)[/math], which is [math](3, \infty)[/math], is then to be considered as the domain of g[math](x)[/math] for considering the range of [math]f(g(x))[/math]. With this domain, the range of [math]g(x)[/math] is [math](9, \infty).[/math]
So, the domain and range of [math]g(f(x))[/math] are [math](1, \infty)[/math] and [math](9, \infty)[/math] respectively.
We work with those subsets. We call them binary relations. There are lots of literature on them.
A function is a relation that for every input, there is exactly one output.
Your composite function is a function.
Functions can be defined in all sorts of ways. Some are defined by an equation, some are defined piecewise, some are just defined by a set of ordered pairs. It really doesn't matter how they're defined as long as for every input there is exactly one output.
Some famous functions that can be defined piecewise are absolute value and the greatest integer function.
OK, let me construct your "composite" function somewhat differently:
[math]f(x)=\displaystyle\frac{1}{2}\left[(x-|x|)x+\sqrt[3]{x}+|\sqrt[3]{x}|\right][/math]
You can verify that it really is the same function, i.e., it maps the values in its domain to exactly the same values as your [math]f(x)[/math] (and it has the same domain, of course.) So then, does this now qualify as one function?
What makes a function a function is not the number of formulas that you use to define it, but the fact that it maps each value in its domain to exactly one value. This is the only thing that matters and makes a function a function; how y
OK, let me construct your "composite" function somewhat differently:
[math]f(x)=\displaystyle\frac{1}{2}\left[(x-|x|)x+\sqrt[3]{x}+|\sqrt[3]{x}|\right][/math]
You can verify that it really is the same function, i.e., it maps the values in its domain to exactly the same values as your [math]f(x)[/math] (and it has the same domain, of course.) So then, does this now qualify as one function?
What makes a function a function is not the number of formulas that you use to define it, but the fact that it maps each value in its domain to exactly one value. This is the only thing that matters and makes a function a function; how you define the function, how many algebraic expressions you use, all that is irrelevant.
Domain mapping is the process of directing visitors from one site to another. It allows you to easily point domains you own to your primary hosting account. You can map subdomains (e.g. http://subsite.example.com) as well as subfolders (e.g. http://example.com/subsite) to a custom domain.
Uniq Web Tech -
Since f.g cannot exist where g does not exist ,its domain would be that of f,excluding where g does not exist.
your f(x)=x²-9
Df=|R
g(x)=√(9-x)
Dg={x:x∈|R, x∈[-3,3]}
So,Dfg=|R ,except what's not in[-3,3]=Dg, only because Df=|R
Since fg(x)=-x² , just looking at the rule of fg we tend to think its donain is the unrestricted |R ,but we need to exclude what we exclude what's not in Dg, everthing outside of [-3,3]
Sounds to me you know absolutely nothing about the subject.
What is a “function”. If you don’t know, look it up. Then answer the question yourself.
What is a “domain” of a function? If you don’t know, look it up. Then answer the question yourself.
What is a “range” of a function? If you don’t know, look it up. Then answer the question yourself.
“How to mapping diagram” sounds like English is 2nd [at least] language for you. If you, you are doing fairly well with it. If it is your native language, then you are not doing fairly well with it.
Let’s say ‘plot’ instead. This works for all such equations
Sounds to me you know absolutely nothing about the subject.
What is a “function”. If you don’t know, look it up. Then answer the question yourself.
What is a “domain” of a function? If you don’t know, look it up. Then answer the question yourself.
What is a “range” of a function? If you don’t know, look it up. Then answer the question yourself.
“How to mapping diagram” sounds like English is 2nd [at least] language for you. If you, you are doing fairly well with it. If it is your native language, then you are not doing fairly well with it.
Let’s say ‘plot’ instead. This works for all such equations - the definition of plotting them:
Stick in a few values of X and do the arithmetic to discover what Y turns up for each.
Find these points on a Cartesian [x,y] graph and put a dot for each one.
Connect the dots. Using appropriately curving ‘lines’ makes it look better.
If you think there is something going on that you don’t see, pick other X values and plot those points, too.
It IS trivia.
Looking at the other [3] answers at this point, I don’t think you’ll learn anything from them. They are written by and for those who understand all of this already.
Is a composite function considered one function?
The function you described in the comment is not what is usually known as a composite function. Your function is defined piecewise. It is still a function.
A composite function is obtained by applying one function to a variable and applying another function (or maybe the same one) to the result. This too, is a function.
As Carter McClung says, a function produces only one output for any input. Your piecewise function does that, so it is a function.
(Historically, it took some time for the modern definition of a function to evolve. It was once more o
Is a composite function considered one function?
The function you described in the comment is not what is usually known as a composite function. Your function is defined piecewise. It is still a function.
A composite function is obtained by applying one function to a variable and applying another function (or maybe the same one) to the result. This too, is a function.
As Carter McClung says, a function produces only one output for any input. Your piecewise function does that, so it is a function.
(Historically, it took some time for the modern definition of a function to evolve. It was once more or less thought of as a formula, which is probably what you are thinking, but we are over that now. The formal definition is a set of ordered pairs (x,y) where there is exactly one pair for each x. The set values of x and y that occur in at least one ordered pair are called the domain and the range.
The advantage of this definition is that it makes no assumption about how the function is defined. You could have a formula, but, for example, with real life measurements you might know that one variable depends on another—a function—but you don’t know exactly what it is, although you might have an approximate formula.)
A composite function is defined as a function of a function. A visualization:
f(g(x))
As you can see, the output of f(x) will be determined in part by the input given- which is, in fact, the output of g(x).
An example
f(x) = x^2 + 2x -3
g(x) = 4x
To solve f(g(x)), we replace all instances of ‘x’ in f(x) with ‘4x’, or in other words g(x).
f(g(x)) = (4x)^2 + 2(4x) -3
= 16x^2 + 8x -3
Simple enough right? One can also input a value of x in g(x) like so:
f(x) = x^2
g(x) = x^3
f(g(2)) = ((2)^3)^2
= (8)^2 = 64
Alternatively:
h(x) = f(g(x)) = (x^3)^2 = x^6
h(2) = (2)^6 = 64
Hope this helps!
The term ‘non-singular’ is usually used in the context of those functions which are linear transformations between vector spaces or their matrix representations, where they mean invertibility of the function.
I may be misunderstanding the question, but I think very nearly every function is composite; that is, any function can be expressed as a composition of functions.
If the question is really about the values of particular functions for particular arguments, that's entirely different: I won't try to address that.
There isn't exactly a notion of “primes” for composition of functions, but there is a notion roughly corresponding to “units”, because bijective functions are invertible, and injective functions have left-inverses.
In more detail:
A function with an empty domain can be considered trivial, an
I may be misunderstanding the question, but I think very nearly every function is composite; that is, any function can be expressed as a composition of functions.
If the question is really about the values of particular functions for particular arguments, that's entirely different: I won't try to address that.
There isn't exactly a notion of “primes” for composition of functions, but there is a notion roughly corresponding to “units”, because bijective functions are invertible, and injective functions have left-inverses.
In more detail:
A function with an empty domain can be considered trivial, and can only be expressed as a composition of functions ending with a function with an empty domain; in this sense it has only trivial expressions as a composition.
Apart from this not very interesting case, you can play around with non-identity bijections until the cows come home, like this:
[math]f = (f \circ g^{-1}) \circ g[/math]
where [math]g[/math] is one of those non-identity bijections on the domain of [math]f[/math].
If you want a composition in terms of non-bijective functions, it depends on [math]f[/math]. If [math]f[/math] is injective, then [math]g[/math] must be injective; re-defining [math]g[/math] (by redefining its codomain to match its image) will leave a suitable bijective [math]g[/math].
If [math]f[/math] is not injective, you can have [math]g[/math] correspondingly not injective. If there are distinct pairs [math]\{a, b\}[/math] and [math]\{c, d\}[/math] of elements of the domain of [math]f[/math], that is, [math]a \neq b[/math], [math]c \neq d [/math]and [math]\{a, b\} \neq \{c, d\}[/math], and [math]f(a) = f(b)[/math], [math]f(c) = f(d)[/math], then you can pick [math]g[/math] so [math]g(a) \neq g(b)[/math] but [math]g(c) = g(d)[/math]. You can't strictly invert [math]g[/math], but you don't need to; you can pick a fake “inverse” [math]h[/math] with [math]h(g(b)) = h(g(a)) = a[/math].
Proceeding in this way to create a [math]g[/math] which maps to the same value some but not all of the pairs that [math]f[/math] maps to the same value, and a corresponding nearly-inverse [math]h[/math] which maps the image values to just one pre-image value, you can leave [math]f[/math] to do the rest.
Now we have
[math]f = (f \circ h) \circ g[/math]
and neither [math](f \circ h)[/math] nor [math]g[/math] is injective, which is reasonably convincingly “not trivial”.
I suspect none of this was what the OP meant!
Yes, as long as everything is defined. This is easiest to show if you work with sequential continuity rather than the standard metric space definition.
The domain of a function is the set of all allowed input vales. For instance, for f(x) = sqrt(x), the domain is x greater than or equal to zero.
The range is the set of all output values on the given domain. For f(x) = sqrt(x), the range is all y greater than or equal to zero.
When a function is always increasing or always decreasing, the domain and range interchange when going from a function to its inverse.
So, for y = x^2, with domain x >= 0, on which the function is always increasing, the range is [0, infinity).
To find the inverse, interchange x and y and solve for y. So, we have x = y^2, whi
The domain of a function is the set of all allowed input vales. For instance, for f(x) = sqrt(x), the domain is x greater than or equal to zero.
The range is the set of all output values on the given domain. For f(x) = sqrt(x), the range is all y greater than or equal to zero.
When a function is always increasing or always decreasing, the domain and range interchange when going from a function to its inverse.
So, for y = x^2, with domain x >= 0, on which the function is always increasing, the range is [0, infinity).
To find the inverse, interchange x and y and solve for y. So, we have x = y^2, which is y = sqrt(x). The domain is [0, infinity) and the range is [0, infinity) too.
Y = sin(x) is increasing on [-pi, pi] and has range [-1, 1]. Y = sin^(-1)(x) has domain. [-1, 1] and range [-pi, pi].
The parameter method
If you have each function given explicitly, say f(x) = x+3 and g(x)=x^2 and you want to find the domain and range of g(f(x)) then the easiest thing to do is form a function in one variable by passing the parameter through. So g(f(x))=g(x+3)=(x+3)^2. Finding the domain and range of this is simple, it's just a normal function. This method applies for higher orders of composition too. However, if you don't have the functions explicitly (we’re not always that lucky) or want to think about it in terms of sets read below.
The set method
Let's say our composite function is g(f(x)).
The parameter method
If you have each function given explicitly, say f(x) = x+3 and g(x)=x^2 and you want to find the domain and range of g(f(x)) then the easiest thing to do is form a function in one variable by passing the parameter through. So g(f(x))=g(x+3)=(x+3)^2. Finding the domain and range of this is simple, it's just a normal function. This method applies for higher orders of composition too. However, if you don't have the functions explicitly (we’re not always that lucky) or want to think about it in terms of sets read below.
The set method
Let's say our composite function is g(f(x)). The easiest way to determine domain and range is to split the functions up.
Start by taking f(x). Determine the range of that function. Take g(x) and determine it's domain. Do this as if they were individual functions - this should be fairly easy.
Now take the intersection of those two sets, call it X. This is the set that links the two functions, it is mapable from the domain of g(f(x)) by carrying out f, and its mapable to the range of g(f(x)) by carrying out g.
The domain of g(f(x)) is the preimage of X in f, or if you prefer f^-1(X)=Y. The range of g(f(x)) is g(X)=Z.
This forms the sequence of equivalences g(f(Y))=g(X)=Z. So Y is the domain and Z is the range. This should be clear by f: Y → X and g: X → Z.
This process can be extended to more highly composite functions such as f(g(h(x))) but requires a bit more moving backwards and forwards. I would start by determining the domain and range of g(h(x)) as above. Then carry out the process of intersecting the domain of f with the range of g(h(x)), call it X. You should then be fairly easily be able to get the total range by calculating f(X)=W. You'll then need to work back to determine the final domain of g(h(x)) that is allowable. To do this you carry out the preimage of g on X, I.e. g^-1(X) = Y. Then carry out the preimage of h on Y, I.e h^-1(Y) = Z. This gives the final domain.
Written more nicely we will have found h: Z → Y, g: Y → X and f: X → W where Z is the domain and W is the range of f(g(h(x))).
I hope this explanation helps. The set methodology requires a bit of thought but after some examples it becomes easier. If anyone has any examples they need help with feel free to comment or request me to answer another question and I'll get round to it.
If g(x) has domain restrictions, that means g cannot take on any values at those restricted values in the domain. Well if g can’t take on any value at those restrictions, then f(g(x)) will be restricted since f cannot take on a value where g(x) is restricted or undefined. A function F (being composite) has nothing to do with “changing” a domain.
We do work with general mappings. They are called relations. Functions are just an important and interesting subset of relations.
A strictly monotonic decreasing (or increasing) function f from a subset of R to R, is obviously one-one, because if a and b are not equal, then a < b or a > b. These
respectively imply f(a) > f(b) or f(a) < f(b),
Aand in either case f(a) and f(b) are not equal. The case of increasing functions is similarly tackled.
Conversely suppose that a function is one-one. Then it need not be either of increasing or decreasing. Consider the function f: R^+ --> R, (R^+ is the set of all non-negative reals) given by
f(x) = x over [2n, 2n+1] for all integers n >/= 0.
and f(x) = (-x) over (2n-1, 2n) for all
intege
A strictly monotonic decreasing (or increasing) function f from a subset of R to R, is obviously one-one, because if a and b are not equal, then a < b or a > b. These
respectively imply f(a) > f(b) or f(a) < f(b),
Aand in either case f(a) and f(b) are not equal. The case of increasing functions is similarly tackled.
Conversely suppose that a function is one-one. Then it need not be either of increasing or decreasing. Consider the function f: R^+ --> R, (R^+ is the set of all non-negative reals) given by
f(x) = x over [2n, 2n+1] for all integers n >/= 0.
and f(x) = (-x) over (2n-1, 2n) for all
integers n >/=0. (For example f(x)=x when
0</=x</=1 and f(x)=(-x) when 1<x<2, etc.)
It may be checked that the function is one-one, but is not monotonic.
On what structure? For groups, it needs to preserve the group operation, for rings it needs to preserve the addition and multiplication, for topological spaces it needs to preserve the topological structure i.e. be continuous, for graphs it needs to preserve the adjacency etc.
In more generality, for a model [math]\mathcal M[/math] of a first order theory [math]T[/math] of signature [math]\sigma[/math], a map [math]\varphi:\mathcal M \to \mathcal N[/math] another model [math]\mathcal N[/math] of [math]T[/math] is a homomorphism if
- it takes [math]\mathcal M[/math]’s interpretation [math]c^\mathcal M[/math] of any constant symbols [math]c[/math] to [math]\mathcal N[/math]’s interpretation [math]c^\mathcal N[/math] of the same symbol, [math]\va[/math]
On what structure? For groups, it needs to preserve the group operation, for rings it needs to preserve the addition and multiplication, for topological spaces it needs to preserve the topological structure i.e. be continuous, for graphs it needs to preserve the adjacency etc.
In more generality, for a model [math]\mathcal M[/math] of a first order theory [math]T[/math] of signature [math]\sigma[/math], a map [math]\varphi:\mathcal M \to \mathcal N[/math] another model [math]\mathcal N[/math] of [math]T[/math] is a homomorphism if
- it takes [math]\mathcal M[/math]’s interpretation [math]c^\mathcal M[/math] of any constant symbols [math]c[/math] to [math]\mathcal N[/math]’s interpretation [math]c^\mathcal N[/math] of the same symbol, [math]\varphi(c^\mathcal M) = c^\mathcal N[/math],
- it preserves atomic statements in [math]\mathcal M[/math] using any of the relations [math]R[/math] of [math]\sigma[/math], [math]R^\mathcal M(t^\mathcal M,s^\mathcal M) \implies R^\mathcal N(\varphi(t^\mathcal M),\varphi(s^\mathcal M))[/math], and
- it preserves images of terms under interpretations of function symbols, [math]\varphi(f^\mathcal M(t^\mathcal M))=f^\mathcal N(\varphi(t^\mathcal N))[/math].
If we have:
[math]h(x)=f(g(x))[/math]
Then:
[math]h^{-1}(x)=g^{-1}(f^{-1}(x))[/math]
You should therefore ‘reverse’ the chain of events.
And study separately for which values [math]x[/math] the above makes sense (in both cases).