Question

f(x)=X(squared)-4x+7 for x>2
g(x)=x-2 for x>2
the function h is such that f=hg and the domain of h is x>0
pleaseee help noww.

Answer 1

So you are trying to figure out h, right?

Answer 2

yes!

Answer 3

Ok, h is such that f = hg,

which means that if you solve for h, you can figure out what is means in terms of f and g.

Given f = hg, what is h? (just in terms of f and g, don't worry about x yet.)

Answer 4

so it becomes h=f/g?
and then i simplify it?

Answer 5

Yes, h = f/g

So, now throw in the expression for f and g.

Answer 6

So what is h in terms of x?

Answer 7

f = x^2-4x+7
g = x-2

Answer 8

yes
i divide f by g??
i think that i have to do something with the domain they gave me
the teacher said that it is more complicated than that

Answer 9

Hmmm, I'm not 100% sure. The only problem I see is that f and g are not defined when 0 < x < 2. Gimme a second to think about that. In the meantime, what is f/g when 2 < x?

Answer 10

if you weree tryin to ask me... i really dont know-.-

Answer 11

plug in the x's for the expressions of f and g to find h

Remember h = f/g

Answer 12

x as a variable or is there any number that represents it?

Answer 13

What is f in terms of x?
What is g in terms of x?

What is h in terms of x?

h = f/g

Answer 14

ahhhh ,okayyyy
thanks a lott:))

Answer 15

So, what is h in terms of x?

Answer 16

(x-2) squared +5 / x-2

Answer 17

do i have to simplify it?

Answer 18

It is simpler if you just leave it in the original form I believe

Answer 19

yeah i guess so
thanks anyway!!!

Answer 20

No problem. In terms of the domain, I'm still not sure but I think h is just undefined from 0 to 2. But you may want to ask someone else about that. Regardless, you have a major part of the problem completed and understand how to solve similar ones.

Answer 21

@Hero , do you have a suggestion on this?

Answer 22

Okay, so do

\[\frac{f}{g} = h , x> 2\]

Answer 23

so what should i substitute x with?

Answer 24

Which you simply divide

\[\frac{x^2 -4x + 7}{x-2}\]

You don't need to substitute x with anything yet

Answer 25

thats it?

Answer 26

Yes, that's it. But @Hero , is h undefined for 0 < x < 2 ?

Answer 27

Also, @Hero can you show me how to do the really spiffy math display you used?

Answer 28

spiffy?

Answer 29

`\[\frac{x^2 -4x + 7}{x-2}\]`

Answer 30

Alternatively, just use equation editor

Answer 31

\[x \ne 2\]

Answer 32

Wow, that's seems complicated lol. I'd rather just type it out like I am :) haha. Thanks tho!

Answer 33

It's not complicated

Answer 34

I understand that x cannot equal 2, but what about x = 1 ? f and g are not defined at x = 1 so therefore h is undefined at x = 1 right?

Answer 35

To produce \(\huge\frac{a}{b}\) place: `\frac{a}{b}` between `\[\]`

Answer 36

According to what was originally expressed, h(x) exists whenever x > 0. You shouldn't try to change the constraints of h(x) based on what the constraints of f(x) and g(x) are. Especially when they explicitly tell you what the constraints are. Don't make it more complicated than it should be.

Answer 37

@teic85

Answer 38

Ok thanks :)

Answer 39

However, the student indicated that their teacher said the domains were important and it wasn't as simple as I made it out to be. I had the same opinion as you on the subject. The domains are what they are. To be honest, I feel like the domains are extraneous information to this problem. @Hero

Answer 40

I agree they are important and recognizing what applies to what is equally important. In that regard, the constraints are clearly specified.

Answer 41

With h(x), the constraints should be (0,2), (2, ∞)