- anonymous

please help im so stuck Given the following functions f(x) and g(x), solve f over g (−5) and select the correct answer below:
f(x) = 2x − 20
g(x) = x − 1

- schrodinger

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- DanJS

f(-5) / g(-5)
that right?

- anonymous

-5 is one of the choices but im not sure

- anonymous

could you explain to me how you got that answer?

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## More answers

- anonymous

−5
5
one sixth
30
the coices

- DanJS

\[\frac{ f(-5) }{ g(-5) } = \frac{ 2(-5)-20 }{ -5 -1 }\]

- DanJS

right , yeah it is f(-5) / g(-5), just evaluate both and put them over each other

- anonymous

ok thank you can you help with few more just trying out openstudy for the first time my friends said it helps

- DanJS

yes, depends on who helps ya.. .. just ask a new thread

- anonymous

you mean like post the question by itself?

- DanJS

screw it, here is fine

- anonymous

ok 1 sec

- anonymous

given the functions f(x)and g(x), solve (f times g)(3) and select the correct answer below
f(x)4x^2+12
g(x)=x-1

- DanJS

f times g, or is it a small circle looking thing

- anonymous

47
-47
96
-96
are choices

- anonymous

its a small cirlce

- DanJS

ok, then it is a compound function f[g(x)]

- anonymous

so how would i set up the equation

- DanJS

nope, it is gonna be multiplication

- DanJS

f(3) * g(3) =

- anonymous

so i would multiply the 2 equation they gave me?

- anonymous

so
(4x^2+12)3 and (x-1)3?

- DanJS

you could do that first, or put in the value for x to start,
[4(3^2) + 12] * (3-1)

- anonymous

hows did you get the value of x

- DanJS

THe problem says [f times g ](3) , x=3

- DanJS

so you just evaluate both f and g at x=3 , then multiply the values together

- anonymous

ooo ok

- anonymous

let me try it

- DanJS

I have time for one more if you want

- DanJS

f(3) * g(3) = [4(3^2) + 12] * (3-1) = 96

- anonymous

how did u get 96 i got 168

- DanJS

[4*9 + 12]* 2

- DanJS

48*2

- anonymous

ok and wat a bout the part when you multiply it by 3-1

- DanJS

that is what happened, the 2 at the end...
f(3) * g(3) = [4(3^2) + 12] * (3-1) = 96

- anonymous

ook

- anonymous

i get it now i messed up omewhere but i see thay why its a 2 at the end cuz 3-1 is 2 ok can u hepl with one more?

- DanJS

yep

- anonymous

Given the following functions f(x) and g(x), solve f[g(6)] and select the correct answer below:
f(x) = 6x + 12
g(x) = x − 8

- anonymous

choices are −96
0
24
48

- DanJS

These compound functions like that, you work from the inside most function first, then outwards carrying the values...

- DanJS

figure out g(6), then use that value in f(x) for x

- anonymous

so x is 6?

- DanJS

yes, anytime you see g(#) or f(#) some number, that is what x is

- anonymous

k so now i plug it in dont tell me ill tell u n u tell if its right

- DanJS

So what is g(6)

- anonymous

is it -2?

- DanJS

no...
First , what is g(6) ?

- anonymous

i did x-8 and plugged 6 for x

- DanJS

right, so g(6) = 6 - 8 = -2

- anonymous

ok so now f

- DanJS

now you use that value you just got in the f function for x

- anonymous

f is 48?

- DanJS

f(-2)

- anonymous

6times 6 plus 12

- DanJS

no, that is the thing with the compound functions, you have to use the value from g(6) = -2
for the value of x in f(x)

- DanJS

Figure the inner function out, use that value in the next function

- anonymous

o ok so i plug the -2 in instead of the 6?

- DanJS

Yes..
f[g(6)] = f(-2)

- anonymous

can you explain how you get that from f(x)=6x+12?

- DanJS

ok, start over from the beginning...

- anonymous

ok

- DanJS

f [ g(6) ]
What that means is instead of having the usual f(x), you have f( another function)
Start with the inner function g(6) and evaluate. g(6) = -2
Now you can replace g(6) with -2 in the original problem

- DanJS

Left to evaluate f(-2)

- anonymous

oo ok

- anonymous

so where do i go from here?

- DanJS

just left to evaluate f(-2) = 6*(-2) + 12

- anonymous

o ok so now u added the 6(-2)+12 wouldnt it be 0

- DanJS

so it is zero
i have to take off, you can fan me , ill prolly be on later

- anonymous

ok how do i fan you?

- anonymous

nvm n did it thanks

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