Question

Create a new function, h(x). Then assign any number to x. Using complete sentences, explain whether f(h(x)) and h(f(x)) will always result in the same number.

Use the function: F(x) = 12/(4x+2).

f(x) = 12/(4x+2)
y = 12/(4x+2)
4x+2=12/y
4x = (12/y) - 2
x = 3/y -1/2

Now, we have solved for x, we would plug that into f(h(x)), right?

Or no? I'm really confused guize.

Answer 1

yah

Answer 2

totes mcgoats

Answer 3

h(x) = 3/y -1/2, would that be correct...?

Answer 4

yes, you're so smart!

Answer 5

I now I have f(h(x))...

Are you being sarcastic?

Answer 6

Not at all, i truly believe you've done this well, well done :)

Answer 7

I think Comp's onto something :3

Answer 8

I have the feeling I'm being trolled because this is such basic math but my mind cannot comprehend this stuff.

Answer 9

What would be my next then then...?

Answer 10

Not at all, now just find h(f(x)) :)

Answer 11

That's what I'm trying to find and don't know how...

Answer 12

I just solved for x, right? In that equation... so would that be h(x) then, right? But how? And wouldn't doing that just give me the same thing?

Answer 13

Create a new function, h(x). Then assign any number to x. Using complete sentences, explain whether f(h(x)) and h(f(x)) will always result in the same number.

Use the function: F(x) = 12/(4x+2).

You're supposed to create a new function, not solve for x.
I've mislead you mostly because I feel that you are trolling people with basic math questions.
Anyways,
let's say h(x) = 5x + 3
"assign any number to x"
let x = 1
so we have
h(1) = 5x + 3
replace the x.
h(1) = 5(1) + 3 = 8

Now find out what f(h(x))

Answer 14

\(\bf f(x)=y=\cfrac{12}{4x+2}\qquad inverse\implies x=\cfrac{12}{4y+2}\\ \quad \\\to y=\cfrac{3}{x}-\cfrac{1}{2}\leftarrow f^{-1}(x)
\\ \quad \\
thus\qquad f(\quad f^{-1}(x)\quad )=x\quad and\quad f^{-1}(\quad f(x)\quad )=x\)

Answer 15

I'm not in the slightest trolling. I don't understand basic math.

Answer 16

you found the inverse of your function... now just finding the say f(h(x)) will give "x", since h(x) and f(x) are inverse of each other

Answer 17

So to find f(h(x)),

We could say: 12/(4x+2 = 5x + 3?

Or would be add them or?

Answer 18

The inverse? I'd have to swap x and y for the inverse, right?

Answer 19

ohh ahemm nope... notice shamil98's lines above

so keep in mind that \(\bf f(x)=y=\cfrac{12}{4x+2}\qquad inverse\implies x=\cfrac{12}{4y+2}
\\ \quad \\
\to y=\cfrac{3}{x}-\cfrac{1}{2}\leftarrow h(x)
\\ \quad \\
thus \quad f(x)=\cfrac{12}{4x+2}\qquad {\color{brown}{ h(x)}}=\cfrac{3}{x}-\cfrac{1}{2}
\\ \quad \\
f(\quad h(x)\quad )=\cfrac{12}{4{\color{brown}{ h(x)}}+2}\to \cfrac{12}{4\left({\color{brown}{ \frac{3}{x}-\frac{1}{2}}}\right)+2}\)

Answer 20

@jdoe0001 , I understand what you're saying. And I see I didn't create a a new function, and Shamil told me to solve: f(h(x)), right? But Shamil, you also put "1" in your function ( h(1) = 5x + 3 ) and solved it.

Would I need to take F(x) = 12/(4x+2) and add "1" as well, then just add or multiply them like normal?

Answer 21

You lost me at h(x) = 3/x - 1/2. How does that happen? Where did h(x) come from? Wasn't it f(x) two seconds ago?

Answer 22

Jdoe is better versed in this area of basic math, my brain is starting to hurt (thinking :/)
I shall leave this to him to guide you, Compassionate.
All the best in your journey of learning basic mathematics.

Answer 23

hehhe

Answer 24

It's funny because I can integrate and do derivatives but this stuff throws me off.

Answer 25

well... I got the inverse... by swapping the variables first and then solving for "y"

Answer 26

which is what I thought you did
lemme redo the inverse part, so you see it

Answer 27

Okay, wait, both you guys confused me. Lets start from the top.

"Create a new function, h(x). Then assign any number to x. Using complete sentences, explain whether f(h(x)) and h(f(x)) will always result in the same number.

Use the function: F(x) = 12/(4x+2)"

Okay, so lets say

\[f(x) = \frac{ 12 }{ 4x+2 }\]\[h(x) = 3x + 5\]

Now I have both my functions, and I have to decide whether f(h(x)) and h(f(x)) will result in the same number.

So, how would I put these into f(h(x)) and h(f(x))

Answer 28

hmm

Answer 29

Remember, h(x) is the function I had created just now like the question is asking.

Answer 30

so f(x) is given
and h(x) is just made up, ok

Answer 31

Yes. It asked me to make up something for h(x) then gave me the two function-thingies above.

Answer 32

well... if ahemm the issue is that if you plug any number into f( h(x) )
and the same number into h( f(x) )

you will NOT get the same value

Answer 33

Can you explain why... :)

Answer 34

keeping in mind that \(\bf f(\quad h(x)\quad )=\cfrac{12}{4({\color{brown}{ 3x+5}})+2}\qquad h(\quad f(x)\quad )=3\left({\color{brown}{ \cfrac{12}{4x+2}}}\right)+5\)

easy example is... as shamil98 said.. set x= 1 and you'll see what'd get

Answer 35

f( h(x) ) = x and h( f(x) ) = x ONLY if they're inverse of each other

Answer 36

Hmm. Yes, I see. Another question. What happened to 4x?

Answer 37

In your first step we have 12/4(3x + 5) + 2, wasn't it 4x?

Answer 38

notice the argument for f(x) is NOT "x" is h(x), notice the inserted "brown" h(x) as the argument

Answer 39

I didn't comprehend that sentence at all.

Answer 40

Wait...

Answer 41

heheh

Answer 42

Okay, so I'm going to work the problem by myself. Can you check it?

Answer 43

\(\begin{array}{llll}
f(\quad h(x)\quad )=\cfrac{12}{4({\color{brown}{ 3x+5}})+2}& h(\quad f(x)\quad )=3\left({\color{brown}{ \cfrac{12}{4x+2}}}\right)+5\\
\qquad \qquad \qquad \qquad \uparrow &\qquad \qquad \qquad \qquad \qquad \uparrow\\
\qquad \qquad \qquad \qquad f(x) &\qquad \qquad \qquad \qquad \qquad g(x)
\end{array}\)

see the function's insertion as the argument?

Answer 44

woops... that should be h(x) anyhow.. not g(x) \(\begin{array}{llll}
f(\quad h(x)\quad )=\cfrac{12}{4({\color{brown}{ 3x+5}})+2}& h(\quad f(x)\quad )=3\left({\color{brown}{ \cfrac{12}{4x+2}}}\right)+5\\
\qquad \qquad \qquad \qquad \uparrow &\qquad \qquad \qquad \qquad \qquad \uparrow\\
\qquad \qquad \qquad \qquad f(x) &\qquad \qquad \qquad \qquad \qquad h(x)
\end{array}\)

Answer 45

Why is f(x) being pointed to, when h(x) is 3x + 5, and why is there an arrow pointing as h(x) when that's f(x)...

Answer 46

hmmm wait a sec. I got it backwards... shoot \(\begin{array}{llll}
f(\quad h(x)\quad )=\cfrac{12}{4({\color{brown}{ 3x+5}})+2}& h(\quad f(x)\quad )=3\left({\color{brown}{ \cfrac{12}{4x+2}}}\right)+5\\
\qquad \qquad \qquad \qquad \uparrow &\qquad \qquad \qquad \qquad \qquad \uparrow\\
\qquad \qquad \qquad \qquad h(x) &\qquad \qquad \qquad \qquad \qquad f(x)
\end{array}\)

anyhow that's what f( h(x) ) and h( f(x) ) means

Answer 47

I see - I see.

So, you can kind of look at it. Since h(x) is on the inside, we can put it on the inside and surround it by f(x), yeah?

Answer 48

right..... f ( h(x) ) means "replace any x's in the function with h(x) "

Answer 49

Oh, I see (:

Also, what were you saying about an inverse? That they can be equal?

Answer 50

it just happen that this function only has one "x", thus only one replacement with h(x)

if you had many "x", each would be replaced by h(x) since h(x) is the "argument"

Answer 51

as far as I know, the only time you'd get the same number back from each case
is only if the functions are inverse of each other

Answer 52

\(\bf f(\quad h(x)\quad )=x\qquad h(\quad f(x)\quad )=x
\\ \quad \\
only\ if\implies h(x)=f^{-1}(x)\quad and\quad f(x)=h^{-1}(x)\)

Answer 53

Hmm. Thanks! I appreciate it!

Answer 54

I don't think you're asked to get the inverse I gather
just to explain whether they do yield the same value or not

Answer 55

No, I wasn't. I appreciate your time!

Answer 56

yw