Question

How do you illustrate the process of writing a function as a composition of simpler functions? Medal goes to the best answer.

Answer 1

I have also searched in my book for examples of how to write a function such as w(x) = 1/t^2+2 as a composition of simpler functions. I am not finding any examples on this whole idea

Answer 2

We are mainly looking for simple parent functions to put together. Like:
f(x) = 1/x
g(x) = x^2
If we take f(g(x)), we put in g(x) for x in the definition of f(x).
f(g(x)) = 1/g(x) = 1/x^2.

Is that w(x) = 1/x^2 + 2, or w(x) = 1/(x^2 + 2) (its all in a denominator)?

Answer 3

I am thrown off by the u(x)

Answer 4

so my functions could be s(x) and t(x) = u(x)

Answer 5

the normal flow I see is f(x) + g(x) = f(g(x))

Answer 6

The original problem is write u(x) = 1/x^2-4 as a composition of simpler functions

Answer 7

Hm.. another question. I just need to know if this is:
\( u(x) = \dfrac{1}{x^2} - 4 \) or \(u(x) = \dfrac{1}{x^2 - 4} \)

Answer 8

Second one

Answer 9

x^2-4 is in the denominator

Answer 10

Okay. So, note that we do have a parent of a reciprocal 1/x function. How can we tell? There is 1 on top and there is a function in the denominator.
\( u(x) = \dfrac{1}{\color{blue}{x^2 - 4}} \)

So what if we called one function g(x) = x^2 - 4, and another one f(x) = 1/x. Can you see how we might combine those to make u(x)?

Answer 11

So you would have 1/x and x^2-4?

Answer 12

Yes, these are two simpler functions. I make up the function names f(x) = 1/x and g(x) = x^2 - 4 just so we can write it more conveniently.

And we can compose the two functions in some way so that we get back u(x).

Answer 13

I understand that part. Does it matter what we call the two function names?

Answer 14

As long as we get back u(x)

Answer 15

I was thinking there had to be some logical order like s(x) and t(x) gives you u(x)

Answer 16

It shouldn't. If you wanted to use that order, you could, but whether we call it f(x) or s(x) it will represent the same thing. :)

Answer 17

This is starting to make sense. :)

Answer 18

Now to find the domain because it is a square root, our domain has to be >= 0

Answer 19

So you would set x^2 -4 as an inequality and solve for x?

Answer 20

Oh? it's a square root?
Like, it was written...
\( u(x) = \dfrac{1}{\sqrt{x^2 - 4}} \) ?

Answer 21

Oh sorry... it was written as a fraction :)

Answer 22

We set the bottom part of the fraction as an inequality... to find the domain?

Answer 23

The bottom part of a fraction can be positive or negative. However, it cannot strictly equal zero.
So we want to solve an equality: Denominator = 0. These will be the points where denominator is 0, thus we would be dividing by zero (bad math).

Answer 24

The way I wrote my answer is f(x) = 1/x, g(x) = x^2-4, u(x) = g(x^2-4) = 1/x / x^2 - 4

Answer 25

We would want to insert g(x) in for x for f(x), to get back to u(x).
Like f(x) = 1/x
f(3) = 1/3
f(x^2) = 1/(x^2)
f(g(x)) = (f o g)(x) = 1/(g(x)) = 1/(x^2 - 4) [g(x) = x^2 - 4]

Answer 26

when you have a composition of functions it just means that instead of the value x you insert a whole other function into the previous one
e.g. if you have \[f(x)=x ^{2}\] and \[g(x)=2x\] f(g(x)) means that instead of the x in the function f you insert the function (g(x)) so f(g(x)) is really f(2x) which is equal to \[(2x)^{2}\]

Answer 27

For the domain, it is going to be x^2 - 4 = 0

Answer 28

Well, the domain is going to be every real number except what you get by solving that equation!

Answer 29

To solve that inequality, you would have x^2 - 4 = 0?

Answer 30

No, because the denominator has to equal 0

Answer 31

The domain of x^2-4 is all real numbers, (-oo, oo)

Answer 32

Is that correct?

Answer 33

To clarify. The domain is actually all values \(x\) so that: \(x^2 - 4 \color{red}\ne 0 \).
If \( x^2 - 4 = 0 \), then we have division by zero which is undefined.

The domain of that function g(x) = x^2 - 4 is all real numbers, yes.
The domain of u(x) = 1/(x^2 - 4) we do need to solve x^2 - 4 =/= 0 to remove values that do not work.

Answer 34

You have x^2 not equal to 4?

Answer 35

isn't the domain really all real numbers except for 2 and -2 because the expression equals 0 in those two cases

Answer 36

Yes, x^2 not equal to 4. Or x not equal to 2 or -2, when we take the square root (and considering both positive and negative roots).

Answer 37

If you want to solve for x, then you take the square root of both sides...

Answer 38

sqrt(x^2) - sqrt(4)

Answer 39

x -2 not equal to 0... x is equal to 2

Answer 40

Oh... 2 and -2

Answer 41

Yes. Taking a square root gives you both a positive and negative branch. Only because:
\( (-2)^2 = 4 \text{ and } 2^2 = 4 \)

Answer 42

Domain: (0, 2] U (-oo, -2)?

Answer 43

I think I messed up somewhere

Answer 44

I think it would be easier to just say the domain is all real values of x where x not equal to 2 or -2.
Interval notation, you need to have the interval from -oo to -2, from -2 to 2, and from 2 to oo.

Answer 45

That is expressed as ||R, x not equal to 2, x not equal to -2

Answer 46

I have used interval notation all the time that I don't know set builder that well

Answer 47

Oh, alright. Domain = { \( x \in \mathbb{R}\) | \(x \ne -2, 2\) } is one way.
But if you are more comfortable with interval notation, we just split up the original interval at x=-2 and x=2.
(-oo, oo) Breaks at x=-2...
(-oo, -2) u (-2, oo) Also at x=+2
(-oo, -2) u (-2, 2) u (2, oo)

Answer 48

Alright. That is what I have for interval notation, but I might start using set notation. I would give you 2 medals if I was able. You have given me a lot of help.

Answer 49

Medals and smart score are just superficial. I am always happy to help! :D

Answer 50

Oh. That is great to hear! I want to be an ambassador and help out people too but I want to demonstrate a higher proficiency in math. :)

Answer 51

Both are nice goals to strive for. And with enough time and effort they are all possible. :)
I know I spent a year or more studying Math on my own time before it just started clicking. It isn't always easy, but it is possible!

Answer 52

By Math, I speak mainly about Calculus I *

Answer 53

I do appreciate the help. Thanks everyone

Answer 54

might I just add that you ought to use square brackets before/after the + and -2 because that's how you denote that they're excluded from the domain whereas round brackets mean that they're included

Answer 55

Oh? I was pretty sure it was the other way around. Round brackets meant exclusion, and square brackets mean inclusion.
[ a, b ] would mean "including endpoints a and b" to me

Answer 56

I will chime in on this one. Brackets mean that it is excluded and ( mean included...

Answer 57

Oh, actually @maida.beganovic is right

Answer 58

[ ] a square bracket if you want to include the end value, or
( ) a round bracket if you don't

Answer 59

That was your post. I think I got it mixed up. You had it right.

Answer 60

http://www.regentsprep.org/regents/math/algebra/ap1/intervalnot.htm
I don't think I've ever seen it the other way with square means exclusion. But if you really wanted to be clear, you could just assert that () means exclusion/inclusion and [] the other. But you have to be consistent, above all.