Inputs and outputs of composite function

[Coco12]

Homework Statement

Can someone describe the input and output method to find the domain and ranges of composite functions to me??

Homework Equations

The Attempt at a Solution

Example: f(g(x))

first you find the domain of g(x)
Then you find the range of g(x)
y= G(x)
that will become the domain of f(x)??

This is the part where I get confused.

 

[haruspex]

Example: f(g(x))

first you find the domain of g(x)
Then you find the range of g(x)
y= G(x)
that will become the domain of f(x)??

No, it should merely be contained within the domain of f(x). f will be defined as a function in its own right independently of g, so its domain can go outside the range of g. Consequently, the range of G could be smaller than the range of f.

 

[Coco12]

Can u explain how it would work then?

 

[haruspex]

Can u explain how it would work then?

The domain of the composite is easy, right? It's just the range that's tricky.
Given g:A→B and f:C→D, the composite f(g(.)) is only meaningful if B\subseteqC. You can define a function $\hat f$:B→D as the restriction of f to B. Then the range of f(g(.)) is the range of $\hat f$.

 

[Coco12]

What is A - B and C-D?

 

[Coco12]

Yup I understand how to do the domain , it is just the range

 

[scurty]

What is A - B and C-D?

$g:A \rightarrow B$ means that if you take any element $x \in A$, $g(x) \in B$; g maps every element in A to some element in B. Basically, you take an element from the set A and put it into the function g and it outputs a solution in the set B. Similarly for C and D for the function f. I think that is what you were asking? (there was no mention of subtraction in haruspex's post)

 

[Coco12]

Is there an example that can be given?

 

[Coco12]

Also , another question let's say if
F(x) = sqrt(x-2)
G(x) = x+3
How do you find the range of (f+g)(x)
Sqrt (x-2) +x+3 ?
Normally the range of sqrt (x+2) will equal all y values greater than or equal to 0. How will adding the x+3 affect it?

 

[scurty]

Is there an example that can be given?

$f:[0,1] \rightarrow [1,2]$ where $f(x) = x + 1$

$g:\mathbb{R} \rightarrow \mathbb{C}$ where $g(x) = xi$

are two examples.

 

[scurty]

Also , another question let's say if
F(x) = sqrt(x-2)
G(x) = x+3
How do you find the range of (f+g)(x)
Sqrt (x-2) +x+3 ?
Normally the range of sqrt (x+2) will equal all y values greater than or equal to 0. How will adding the x+3 affect it?

I believe Mark44 answered a similar question in this topic: https://www.physicsforums.com/showthread.php?t=728375

Consider $\sqrt{x-2}$ by itself. The smallest value obtained is 0 when $x = 2$ and it tends towards $\infty$ as $x \rightarrow \infty$. Now consider $\sqrt{x-2} + x + 3$. This is only valid, again, for $x \geq 2$. What is the value when $x = 2$? How does this relate to the smallest value obtained from just the square root?

Alternatively, like Mark44 suggested, you can also look on a graph.

 

[haruspex]

Also , another question let's say if
F(x) = sqrt(x-2)
G(x) = x+3
How do you find the range of (f+g)(x)
Sqrt (x-2) +x+3 ?
Normally the range of sqrt (x+2) will equal all y values greater than or equal to 0. How will adding the x+3 affect it?

In general, you need to find all the extrema. Each extremum may be interior to an open set in the domain or at a domain boundary. In the first case, you look for the derivative being zero (if differentiable). In the second, you have to identify the domain boundaries that are contained in the domain. In your example x=2 is the only such. For a function of two variables it could be a surface.
To complete the solution, you need to find the value at each extremum and whether it is a local maximum or minimum.

 

[Coco12]

Let's say g(f(x))
G(x)=x+1
F(x)= 4 ^x

I got 4^x +1

That would be an exponential graph with an horizontal asymptote at 1
The range would be y greater than 1
The book says y greater AnD equal to one .. Why is that so?

 

[Coco12]

I believe Mark44 answered a similar question in this topic: https://www.physicsforums.com/showthread.php?t=728375

Consider $\sqrt{x-2}$ by itself. The smallest value obtained is 0 when $x = 2$ and it tends towards $\infty$ as $x \rightarrow \infty$. Now consider $\sqrt{x-2} + x + 3$. This is only valid, again, for $x \geq 2$. What is the value when $x = 2$? How does this relate to the smallest value obtained from just the square root?

Alternatively, like Mark44 suggested, you can also look on a graph.

When u plug in 2 it would give you 5 as opposed to 0. What next?

 

[scurty]

Let's say g(f(x))
G(x)=x+1
F(x)= 4 ^x

I got 4^x +1

That would be an exponential graph with an horizontal asymptote at 1
The range would be y greater than 1
The book says y greater AnD equal to one .. Why is that so?

I would say the book is wrong and you are right, unless for some odd reason you are adjoining $-\infty$ to the real number line. I would double check with your teacher just to make sure.

 

[Coco12]

When considering range do we only look at the range of the difference of functions or have to consider range of both or just the final function

 

[scurty]

When u plug in 2 it would give you 5 as opposed to 0. What next?

In Calculus we would call this function strictly increasing, because as x increase, y increases, no matter how small the increment of x. For now you can just refer to a graph and see how it isalways increasing.

Because it is always increasing, the upper limit is infinity like before. However, it has a lower limit, because the domain is restricted to $[2,\infty)$. You already found the two values, 0 and 5. Adding the x + 3 term raised the range from $[0,\infty) \text{ to } [5,\infty)$.

 

[Coco12]

In Calculus we would call this function strictly increasing, because as x increase, y increases, no matter how small the increment of x. For now you can just refer to a graph and see how it isalways increasing.

Because it is always increasing, the upper limit is infinity like before. However, it has a lower limit, because the domain is restricted to $[2,\infty)$. You already found the two values, 0 and 5. Adding the x + 3 term raised the range from $[0,\infty) \text{ to } [5,\infty)$.

So the range is everything greater than or equal to 5? How do u know whether it is not equal to, less than or less than and equal to etc.

 

[scurty]

When considering range do we only look at the range of the difference of functions or have to consider range of both or just the final function

The final range is only important. Consider the example:

$f(x) = \sqrt{x}, g(x) = 2\sqrt{x}, \, f(x) - g(x)$

f(x) and g(x) have ranges of $[0,\infty)$, but f(x) - g(x) has range $(-\infty,0]$. Also, consider the case where f(x) = g(x). Their subtraction would result in a range of just 0.

 

[scurty]

So the range is everything greater than or equal to 5? How do u know whether it is not equal to, less than or less than and equal to etc.

Yep! As for why: assuming you haven't taken Calculus yet, graphing is the best you can do. When you hit Calculus you will learn techniques to prove that 5, in this case, is the lower value for the range.

I hope that in this case it is clear as to why 5 is the lowest value in the range, but other functions may be more complicated than this one.

 

[Coco12]

The final range is only important. Consider the example:

$f(x) = \sqrt{x}, g(x) = 2\sqrt{x}, \, f(x) - g(x)$

f(x) and g(x) have ranges of $[0,\infty)$, but f(x) - g(x) has range $(-\infty,0]$. Also, consider the case where f(x) = g(x). Their subtraction would result in a range of just 0.

Ok look at this example:

H(x)= G(x) / f(x)
G(x)= 1/(x+4)
F(x) = 1/(x^2-16)
H(x)= (x-4)(x+4)/(x+4)

I put : y cannot equal to -8
The book says y cannot equal to -8 and 0
Why the zero?

 

[scurty]

Ok look at this example:

H(x)= G(x) / f(x)
G(x)= 1/(x+4)
F(x) = 1/(x^2-16)
H(x)= (x-4)(x+4)/(x+4)

I put : y cannot equal to -8
The book says y cannot equal to -8 and 0
Why the zero?

I was wrong in my previous post, sorry. You do need to consider the original functions, because if the domain is restricted, that will affect the range. In this case -4 is not allowed which prevents -8 from being in the range.

As for why 0 is not in the range, the same principle applies. In f(x), you have $x^2-16$ in the denominator. That factors into (x-4)(x+4). This means x = -4,4 are NOT in the domain of f(x). So, H(4) is undefined because $H(4) = \frac{G(4)}{F(4)}$, but F(4) is undefined. H(x) simplifies to just x-4 so if you plug in 4 and -4 you get 0 and -8; that's where the values came from.

 

[Coco12]

I was wrong in my previous post, sorry. You do need to consider the original functions, because if the domain is restricted, that will affect the range. In this case -4 is not allowed which prevents -8 from being in the range.

As for why 0 is not in the range, the same principle applies. In f(x), you have $x^2-16$ in the denominator. That factors into (x-4)(x+4). This means x = -4,4 are NOT in the domain of f(x). So, H(4) is undefined because $H(4) = \frac{G(4)}{F(4)}$, but F(4) is undefined. H(x) simplifies to just x-4 so if you plug in 4 and -4 you get 0 and -8; that's where the values came from.

So u plug in 4 and -4 into the equation to get the range?

What about if the final equation was 1/(x-1) +sqrt(x)
The domain is x greater than or equal to 0 and x cannot equal 1
. Note this is a sum of functions
How does that plug into the range??

 

[scurty]

So u plug in 4 and -4 into the equation to get the range?

What about if the final equation was 1/(x-1) +sqrt(x)
The domain is x greater than or equal to 0 and x cannot equal 1
. Note this is a sum of functions
How does that plug into the range??

No, the 4 and -4 were undefined values for the domain. To see what values in the range these corresponded to, I plugged them into H and then I discarded those results.

You can't plug in 1 for the new function, because x-1 wasn't canceled out in the function.

For the other function, if you simplified it, it would become: $H(x) = G(x)/F(x) = \frac{x^2-16}{x+4} = \frac{(x+4)(x-4)}{x+4} = x-4$. Because the x+4 canceled out, in the new equation you can get a value for x = -4. However, this is invalid because -4 was excluded from the domain on F(x) (and whenever you cancel terms out, you exclude those values of x that make the term 0 from the final answer).You don't find ranges by plugging in one or two numbers. Use the graphing method for now.

 

[Coco12]

The graphing method is inaccurate isn't it?

 

[scurty]

A graph is inaccurate, yes, but it allows you to see the big picture.

To focus on the smaller details, you need to consider values that make the function undefined, those are excluded from the domain, and how they affect the range. There are so many different types of examples that there is no single way of going about to do this. In square roots, you can't have negatives under the radical. In fractions you can't have a 0 in the denominator. In natural logs, you can only have positive numbers in the domain. There's lots of examples and these all affect the range. Because you can't use Calculus, you aren't going to get complicated functions, a graph should be sufficient. How were you taught in class?

 

[Coco12]

We were taught to plug in domain into ranges.
Then graphing. We did not spent as much of time discussing domains and ranges therefore it's not that clear.
Is this a good general idea?
If the domain cannot equal something , you plug into the equation to get the range values
If the equation is something that can be easily visualized then you can just graph and determine range values

 

[scurty]

We were taught to plug in domain into ranges.
If the domain cannot equal something , you plug into the equation to get the range values

I don't understand what the first sentence means.

As for the second, that is only the case where you eliminate a solution from the bottom of a fraction. Like in the case above where you canceled out the x+4 terms. In any other case, if you try to plug in an answer that is not in the domain, you can't get an answer.

 

[Coco12]

Meaning that if the domain is:
For example x greater than or equal to 2 and x less than or equal to 3.
You plug in x into the equation to find out what numbers y is greater than or equal to and same for the 3

 

[scurty]

Okay, that makes sense now. You plug the x values into the equation, not the range.