Question

If I was to create my own function, and then assign any number to x.
How would I explain whether f(h(x)) and h(f(x)) will always result in the same number?

Answer 1

@ParthKohli could you help?

Answer 2

Hi, have you heard of an inverse function?

Answer 3

would it help if I went ahead and made a function?

Answer 4

i have.

Answer 5

Yes. Two inverse functions are just like that.

Answer 6

you're saying f(h(x)) and h(f(x)) are the inverse of each other?

Answer 7

Let's say my function was this----> F(x)=7x-2

Answer 8

you'll get your medals, I appreciate the help.

Answer 9

Yes, that's how I mean it. For example, multiplying and dividing by the same number.

Answer 10

Eh, no medals needed... x/2 and 2x are inverse functions.

Answer 11

wow, you the man.

Answer 12

?

Answer 13

okay I see what you mean.

Answer 14

I'd need to create an inverse function? In order to show this.

Answer 15

Yes, any two functions that are inverses of each other would work.

Answer 16

how would that show whether f(h(x)) and h(f(x)) will always result in the same number? @ParthKohli

Answer 17

OK, let's show it.

Let's say that
h(x) = x * 1/2
and
f(x) = 2 * x

What is f(h(x))?

Answer 18

I'd have to solve for X, solve each one individually, and then do the same for h(f(x)) and Voila, that's my answer?

Answer 19

@ParthKohli

Answer 20

Hello there, mate~
Mind if I watch you tackle this minor fiasco? ^_^

Answer 21

Yes, that is right.

Answer 22

My fiasco, is your fiasco...we've just become brothers @terenzreignz

Answer 23

Can you calculate f(h(x))?

Answer 24

@ParthKohli you rock, thanks. :)

Answer 25

I don't know how to calculate F(h(x)) :/

Answer 26

Eh, not necessarily... but thanks!

You know what f(h(x)) means, right?

Answer 27

Brothers, are we? Well then, for starters, it's TJ :)

Perhaps we might need a little review of what functions are and their not-so complex complexities?

Answer 28

We're doing the thing "h" to f(x).

h(x) = 1/2 * x

Doing the thing "h" means that you take something and halve it.

Answer 29

So, you are going to take f(x) and halve it.

f(x) = 2 * x

right?

What is the half of 2 * x?

Answer 30

In \(h(\color{blue}{f(x}))\), that is.

Answer 31

The complexity of their level of complex reaches into layers and layers of complexity until you find yourself, much like the panda that learned Kung-Fu, in a void with no escape...lol I'll take all the help I can get @terenzreignz

Answer 32

would it be x/2? O.o @ParthKohli

Answer 33

\[h(f(x))\]\[= h(\color{blue}{2\times x})\]\[= \dfrac{1}{2}\times \color{blue}{2 \times x}\]

Answer 34

arhhhg, sorry man...Math is dead to me. :/ I suck at it

Answer 35

that's the final product? O.o

Answer 36

Nope! What is 1/2 * 2 * x?\[\left(\dfrac{1}{2}\times 2\right) \times x\]

Answer 37

you see to me that looks like a bunch of mumbo-jumbo...I would say
1 * x

Answer 38

Yes! That is right. 1 * x is just x, right?

Answer 39

because .5, or half of 1 multiplied by 2, is...1

Answer 40

yah, it would just be x.

Answer 41

Yup, you got it.

How about f(h(x))? Since the question is looking for both h(f(x)) and f(h(x))...

Answer 42

ok so the final product for hfx would be just X?

Answer 43

and I would say...fhx would be 2x

Answer 44

The final product of h(f(x)) is just x, yes.

How can f(h(x)) be 2x...? Ideally, they should both be x as they result in the same number. :P

Answer 45

May I give it a shot? Watching is boring. :3

Answer 46

lol it's all you man

Answer 47

are you solving for F(x)=7x-2? that function :$

Answer 48

All right, Jack.. first, no tagging me all the time, I'm already on this question, I got a name, it's TJ (not really, LOL), so use it.

Now, to more important matters... a function is like a process... it takes an input, usually called x, and sort of... *does something* to it. It can be anything really, from doubling...
\[\Large f(\color{red}x) = 2\color{red}x\]
To squaring...
\[\Large f(\color{red}x)= \color{red}x^2\]
Or even just raising by 4...
\[\Large f(\color{red}x) = \color{red}x+4\]

Catch me so far?

Answer 49

so far TJ, I follow.

Answer 50

Good. Now, x is technically just a placeholder... see that first function, the 'doubling' function?

We can replace x with virtually anything (no wisecrack comments please :3 )
And the function will have a value...or output.

Say, \[\Large f(\color{red}2) = 2(\color{red}2)=4\]
Makes sense, yeah? Doubling 2 gives us 4.

Still catching me so far?

Answer 51

oooooh

Answer 52

it clicked... -.-

Answer 53

that's the point of the question!

Answer 54

that if the answer always ends up the same, which it will, then you can literally put anything in X.

Answer 55

It makes sense now..

Answer 56

Focus. Try not to jump to conclusions. Arriving at the correct conclusion through the wrong logic occurs quite often in Maths, so stay on track.

Now, when I said we can replace x with *anything* I mean even other functions.

Suppose we have two functions. The doubling function\[\Large f(\color{red}x)=2\color{red}x\] and the "raise-by-four" function:\[\Large g(\color{red}x)= \color{red}x+4\]

Still with me?

Answer 57

yah.

Answer 58

And just to illustrate, with the doubling function, the answer will NOT always be the same regardless of what we replace x with. Try doubling 2 and 3. You get 4 and 6, which aren't the same. So.. now...

Answer 59

Enter a term. It's called "Function composition" It's when you apply a function to x, and then apply another function, NOT to x, but to the result of the first function.


^That may have been a lot to take in, so read through it, devour it, digest it.

And then I'll illustrate an example.

Answer 60

Digested *yum*

Answer 61

Right. So what if I ask you for a function that FIRST doubles a number, and then adds four to it?

f(x) was our doubling function, and g(x) was our "raise-by-four" function.

The composition function is set up as follows:
First you double:\[\Large f(\color{red}x)\]... and then you raise it by four:
\[\Large g\left(\color{red}{f(x)}\right)\]

^ A bit tricky to get, so again, read thoroughly.

Answer 62

Ready? What is this going to look like, I wonder...

Answer 63

I'm honestly a bit confused at this point..

Answer 64

Okay, it's good that you tell me in advance.
If I ask you to double 5 and then add 4 to it, what would be the answer? ^_^

Answer 65

14

Answer 66

Great. So you can actually (albeit not consciously, apparently) compose functions.

Now to the more technical stuffs...

f(x) is the doubling function, so we apply it first, right?

Answer 67

I would say so.

Answer 68

Right, and whatever the output (represented by f(x) itself) we apply g(x) TO it.
Still following? ^_^

Answer 69

so far lol I follow

Answer 70

you just left </3...

Answer 71

When you do return...
So, basically, we apply the function g to f(x).
Like so:
\[\Large g(\color{red}{f(x)})\]
Like a mutant function of sorts.

Now, I'd just like to recall that \[\Large g(\color{red}x)=\color{red}x+4\]
Notice that whatever is in the parentheses beside g, it simply takes the place of all the x's in the function.

Like so:
\[\Large g(\color{red}y)=\color{red}y+4\]\[\Large g(\color{red}{a+b})=\color{red}{a+b}+4 \]

And so on... still catching me?

Answer 72

still catching

Answer 73

I really have to solve this problem soon..

Answer 74

So, what about \[\Large g(\color{red}{f(x)}) \]
?

By that logic, you'd probably think that it should equal\[\Large g(\color{red}{f(x)})= \color{red}{f(x)} +4\]
right?

Answer 75

I'd say so :)

Answer 76

And you'd be right.
And remember that f(x) is simply 2x.
So replace again, accordingly, and you get
\[\Large g(f(\color{red}x))=2\color{red}x + 4 \]

Answer 77

so g=4 and f(x)=2x

Answer 78

hence, why they're equal :)

Answer 79

makes more and more sense

Answer 80

No....
g(x) = x+4
f(x) = 2x

When you have g(f(x)), simply replace the x in g(x) with f(x), like so:

g(f(x)) = f(x) + 4

And remember that f(x) was originally 2x, so

g(f(x)) = 2x + 4

Answer 81

okay

Answer 82

so with that in mind...could you help me solve the question? O.o

Answer 83

I have to create my own function, assign any number to x, explain whether f(h(x)) and h(f(x)) will always result in the same number. ...which they will

Answer 84

I just need help explaining how lol

Answer 85

Yes, well, there are certain pairs of functions that have a unique property... that is, they cancel each other out.

These functions are called inverses.

Think about it... it's like an 'undo'

Like when doubling, to undo it, you halve.
When squaring, to undo it, you get a square root.
When adding four, to undo it, simply take away four.

Get the picture?

Answer 86

OH, is THAT the question?
I thought you were asked to find a pair of functions where f(h(x)) is always equal to h(f(x))

-_-

Answer 87

lol nope, that's the question I posed in the beggining.

Answer 88

you have helped though, in my understanding of functions.

Answer 89

Well then, that simplifies things somewhat. You still have to learn about compositions, though. Having said that, have you made up your own function?

Answer 90

Are you already given a function?

Answer 91

I'll make own now... f(x)=4x-1

Answer 92

is that okay for a function?

Answer 93

Of course. Although, from the looks of it, you have to make up another one.
Unless you were already given another function...

h(x)?

Answer 94

sooo... h(x)=4x-1?

Answer 95

it says to create your own function..

Answer 96

It does... but does it say just one?
Perhaps you need to make up two functions.
sure they can be the same function as you have just done now, but that's boring, think of something else.

(NOTE, this is IF you weren't given a specific function for h or f)

Answer 97

just one