Question

f(x) = 3x + 6, g(x) = 2x^2

Find (fg)(x).

Answer 1

@freckles

Answer 2

i want to be sure I read the problem correctly

is that
\[(f \circ g )(x)\]
or
is that
\[ (fg)(x)\]

it is probably the last since it is was you wrote
but sometimes if you copy and paste from something else it omits certain symbols

Answer 3

its the last one

Answer 4

so (fg)(x)

Answer 5

ok so we are multiplying then

Answer 6

for this you just need to know distributive property and law of exponents really

Answer 7

f(x) = 3x + 6, g(x) = 2x^2

(fg)(x) is just f(x)*g(x)

Answer 8

that means you are doing (3x+6)(2x^2)

Answer 9

\[(3x+6)(2x^2)\]

Answer 10

they will most likely want you to multiply that out

Answer 11

so it would be 6x^3+12x^2?

Answer 12

bingo!

Answer 13

ok, thanks! there is a more complicated one though that i probably should've asked instead of that one

Answer 14

its this: f(x) = Square root of quantity three x plus seven. , g(x) = Square root of quantity three x minus seven.

Find (f + g)(x).

Answer 15

so (f+g)(x) just means you are doing f(x)+g(x)

Answer 16

ok, so it would just be the two added together and thats all?

Answer 17

yep

Answer 18

and i forgot though how do i add square roots again?..

Answer 19

is this right?
\[f(x)=\sqrt{3x+7} \text{ and } g(x)=\sqrt{3x-7 }\]

Answer 20

yes

Answer 21

would it be sqrroot 6x^2

Answer 22

nope they aren't like terms

Answer 23

ok, so what would i do?

Answer 24

so just add them
you can't do nothing else
no like terms
can't be simplified

Answer 25

if you had \[f(x)=\sqrt{3x}+7 \text{ and } g(x)=\sqrt{3x}-7\]
this would be a different story

Answer 26

oh ok, so it would just be the two sqrroots with an addition sign in between them for the answer?

Answer 27

(f+g)(x) can be simplified in this case

Answer 28

yep

Answer 29

oh wow. Ok, thanks again. I am doing a study guide and needed MAJOR review. I will need your help on a few more questions as well as I need to know the process of how to get the questions . Is it ok if i keep tagging you in questions?

Answer 30

and for the other problem I mentioned
\[\text{ if } f(x)=\sqrt{3x}+7 \text{ and } g(x)=\sqrt{3x}-7 \\ \text{ then } (f+g)(x)=f(x)+g(x) \\ (f+g)(x)=[\sqrt{3x}+7]+[\sqrt{3x}-7] \\ (f+g)(x)=\sqrt{3x}+\sqrt{3x}+7-7 \\ (f+g)(x)=2 \sqrt{3x}+0 \\ (f+g)(x)=2 \sqrt{3x}\]

Answer 31

i need to take a short break

Answer 32

ok, thanks!