Help me with some questions?
Given f(x) = 6x2 – x – 12 and g(x) = 2x – 3, find the function (fg)(x).
Evaluate the composite function f(g(x)) for x = 31.

- anonymous

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- anonymous

Much gratitude to whoever helps me. I just can't get these. I hate alg 2

- terenzreignz

f + g ?
You just add f(x) and g(x)

- anonymous

would the answer just be x^2? for that one?

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## More answers

- terenzreignz

As a matter of fact, yes :D

- terenzreignz

now the composite function f(g(x))
is a little tricky
What you do is replace all instances of x in f(x) with g(x)

- terenzreignz

\[\large f(\color{orange}x) =2 - \color{orange}x^2\]\[\large \color{red}{g(x)=3x+10}\]
\[\Large f[\color{red}{g(x)}]=2-[\color{red}{g(x)}]^2\]

- anonymous

And then I just solve that?^

- terenzreignz

replace g(x) and then solve from there, yes :D

- anonymous

okay I get x-5. Is that correct?

- terenzreignz

I don't think so. Remember
\[\large g(x) = 3x+10\]

- anonymous

hmmm, is it 6x+20?

- terenzreignz

You're guessing. Replace the g(x) here
\[\Large f[\color{red}{g(x)}]=2-[\color{red}{g(x)}]^2\]
with 3x + 10 like so...
\[\Large f[\color{red}{g(x)}]=2-[\color{red}{3x+10}]^2\]

- terenzreignz

and carry on from there...

- anonymous

No i'm not. Do I just combine? Coz what I did was multiplied within the parentheses.

- terenzreignz

First, you evaluate
\[\Large [3x+10]^2\]

- anonymous

6x+20 right?

- anonymous

9x sorry

- anonymous

ugh wait 9x+100

- anonymous

@terenzreignz

- terenzreignz

Could you review the FOIL method or the method of squaring a binomial?
Or, as a quick reference, look at this...
\[\Large (\color{red}a+\color{blue}b)^2 = \color{red}a^2 + 2\color{red}a\color{blue}b+\color{blue}b^2\]

- terenzreignz

By that logic, what is
\[\Large (\color{red}{3x}+ \color{blue}{10})^2= \qquad?\]

- anonymous

ahhh okay 3^2+2(3)(10)+10^2 like that?

- terenzreignz

Do remember the x...
your 'a' in this case is not simply 3 but 3x

- anonymous

okay 3x^2+2(3x)(10)+10^2

- anonymous

then combine?

- terenzreignz

\[\large3x^2\qquad \color{red}?\]

- anonymous

I don't understand what you're asking

- terenzreignz

Why is it 3x^2 ?
What is the square of 3x ?

- anonymous

oh 9x^2

- terenzreignz

better. Okay, so simplify?

- anonymous

okay 3x^2+2(3x)(10)+10^2= 9x+6x+5+100

- anonymous

or I mean 105?

- terenzreignz

Why do you have 3x^2 again? -.-

- anonymous

no I changed it to 9x+6x+105

- terenzreignz

let's start again...
\[\Large (3x)^2+2(3x)(10)+10^2\]
And simplify this one at a time, please.

- anonymous

okay so (3x)^2=9x

- terenzreignz

9x^2 !!!

- anonymous

where are you getting 9x^2 from? but anyways 81x

- terenzreignz

No... I mean, you square both the 3 and the x -.-
\[\Large (3x)^2=\color{red}{9x^2}\]

- anonymous

oh okay. Got it

- anonymous

9x^2+6x+5+100x^2 better?

- terenzreignz

Nope. Now what about this part (in red)?
It's just multiplication.
\[\Large9x^2 + \color{red}{2(3x)(10)}+10^2\]

- anonymous

sorry 6x+20 right?

- terenzreignz

wrong.
what's 2(3x)(10)

- anonymous

60x

- terenzreignz

right. now this.
\[\Large 9x^2 + 60x +\color{red}{10^2}\]

- anonymous

100

- terenzreignz

\[\Large 9x^2 +60x + 100\]
And this is what we put in this place...
\[\Large f[g(x)]= 2 - \color{red}{[g(x)]^2}\]

- anonymous

omg sorry for taking up so much of your time

- terenzreignz

you won't be for longer, lol, I have to get to class in a few minutes

- anonymous

ah okay thanks anyways!

- terenzreignz

I'm sure there are other people on OS who can help you out, but unfortunately, for the moment, I have to go :D
--------------------------------------------
Terence out

- anonymous

Still need help with this problem?

- anonymous

yes!

- anonymous

the person answered a different question before this

- anonymous

this problem?
Given f(x) = 6x2 – x – 12 and g(x) = 2x – 3, find the function (fg)(x).
Evaluate the composite function f(g(x)) for x = 31.

- anonymous

before we start though, do you know what g(2) = ?

- anonymous

no they didn't answer that. and no..

- anonymous

g(2) is what you get when you plug 2 for x in g(x)

- anonymous

g(2) = 2(2) -3 = 1

- anonymous

since g(x) = 2x - 3

- anonymous

so f(g(x)) is the same principle. we plug in g(x) in every x in f(x), same way that g(2) is plugging 2 in every x of g(x)

- anonymous

so f(g(x)) = 6[g(x)]^2 - g(x) - 12
do you follow so far?

- anonymous

Ohh okay

- anonymous

so we have:
\[f(g(x)) = 6(2x - 3)^2 - (2x -3) -12\]
knowing the shortcut, by habit, that:
\[(a+b)^2 = a^2 + 2ab + b^2\] i'll save my self a small step and a small mess to expand.
\[f(g(x)) = 6(4x^2 - 12x + 9) - 2x + 3 - 12 = 24x^2 -74x + 45\]
that's (fg)(x). now they ways (fg)(31) which is equivalent to f(g(31)) [diffrent notation]
plug in 31 for every x in f(g(x))

- anonymous

now they want**

- anonymous

:/ oof this..problem looks intimidating

- anonymous

it's easier than it seems. and with a bit of practice they will become very easy and second nature.
i have may have complicated things with the "shortcut". expanding with foil is as good.
don't let math intimidate you, it's your friend :)
and we're here to help you love it!

- anonymous

okay so with every x in there I replace is with a 31? @Euler271

- anonymous

yup

- anonymous

okay 6(4(31)^2-12(31)+9)-2(31)+3-12=24(31)^2+74(31)+45

- anonymous

@Euler271

- anonymous

ya they would be the same thing

- anonymous

Um for the first part I get 20815... @Euler271

- anonymous

i get 25403

- anonymous

oh okay the 2nd part I get 25403 too

- anonymous

@Euler271

- anonymous

cool ^_^
thats the final answer

- anonymous

Oh! @euler. those were both two different problems though...

- anonymous

@euler this was for the 2nd problem

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