Question

if h(x)=f(x)g(x) and f(x)=2x+5, determine g(x)

a) h(x)=3x-1/x+7

b h(x)=3x-1/the square root of x+6

Answer 1

a)\[h(x) = \frac{ 3x-1 }{ x+7 }\]

Answer 2

and b) \[h(x)=\frac{ 3x-1 }{ \sqrt{x+6} }\]

Answer 3

You can start by writing after the equation in part (a) for h(x) that it also equals f(x)g(x) which equals (2x+5)g(x) according to the given

Answer 4

\[h(x) = \frac{ 3x-1 }{ x+7 } = (2x+5)g(z)\]

Answer 5

Then divide both sides by (2x+5) to isolate g(x) (sorry, I accidentally put g(z) in the last equation)

Answer 6

Then simplify the fraction expression if you can

Answer 7

Let me know if this makes sense... :)

Answer 8

so would it be like \[g(x)=\frac{( \frac{ 3x-1 }{ x+7 }) }{ 2x+5 }\] and then simplify it?

Answer 9

yes :) You can start simplifying by "moving" that (x+7) term down to the bottom, like:
g(x) = (3x-1) / [ (x+7)(2x+5) ]

Answer 10

But that's the idea... they give you h(x) = f(x)g(x) and they give you f(x) and h(x) as expressions, so you just substitute those expression in where they belong, and you are left with g(x) and a bunch of x terms and constants to simplify to obtain g(x).

Answer 11

In basic algebra, you might substitute in values for variables to solve for one variable. Same idea here, but you sub in function/expressions for the known functions to solve for the function you don't know, in this case, g(x)

Answer 12

ok

Answer 13

i cant really get how to simplify it past \[g(x)=(3x-1)/3^{2}+19x+35\]

Answer 14

where you have 3^2 on the bottom, it's actually x^2, but yeah, I see what you mean... I'm not seeing a simplification yet.

Answer 15

That may be ok... nothing says g(x) must be a simple expression. Although usually if things get too messy, it's good to look for an simple sign error or something.

Answer 16

k, so would i do the second one the same way to?

Answer 17

Sorry, stepped away... yes, the second one is the same exact idea, and it too looks messy with likely no way to simplify that square root term... don't worry that the g(x) expressions don't "look nice", just simplify what you can. They gave you messy h(x) expressions... that's what's causing it.

Answer 18

ok, there was another one to, it had all of the same h(x) things and it was h(x)=6x+15

Answer 19

so then i got 6x+15=(2x+5)(g(x))

Answer 20

Remember, h(x) = f(x)g(x) according to the given. So, 6x+15 = (2x+5)g(x). Solve for g(x)
:) You beat me to it :)

Answer 21

Note that you finally (!!) can factor the expression to simplify it...

Answer 22

and then \[\frac{ 6x+15 }{ 2x+5 } = g(x)\]

Answer 23

Always look at the denominator to see if it might be an easy factor of the numerator

Answer 24

factor out 3 on top --> 3(2x +5)
and cancel with the denominator term
leaving g(x) = 3

Answer 25

Then say "duh" because when you look at h(x), it's just 3 times f(x), so g(x) = 3

Answer 26

I didn't see that at first

Answer 27

haha thanks, ya i was like ooooh

Answer 28

Nice!

Answer 29

ok, i get how to do this now, thanks allot :D

Answer 30

Glad to help!!

Answer 31

glad for the help :P lol