f(x) = 3x + 6, g(x)… - QuestionCove
OpenStudy (iwanttogotostanford):

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

1 year ago
OpenStudy (iwanttogotostanford):

@freckles

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (iwanttogotostanford):

its the last one

1 year ago
OpenStudy (iwanttogotostanford):

so (fg)(x)

1 year ago
OpenStudy (freckles):

ok so we are multiplying then

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (freckles):

f(x) = 3x + 6, g(x) = 2x^2 (fg)(x) is just f(x)*g(x)

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (freckles):

$(3x+6)(2x^2)$

1 year ago
OpenStudy (freckles):

they will most likely want you to multiply that out

1 year ago
OpenStudy (iwanttogotostanford):

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

1 year ago
OpenStudy (freckles):

bingo!

1 year ago
OpenStudy (iwanttogotostanford):

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

1 year ago
OpenStudy (iwanttogotostanford):

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).

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (iwanttogotostanford):

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

1 year ago
OpenStudy (freckles):

yep

1 year ago
OpenStudy (iwanttogotostanford):

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

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (iwanttogotostanford):

yes

1 year ago
OpenStudy (iwanttogotostanford):

would it be sqrroot 6x^2

1 year ago
OpenStudy (freckles):

nope they aren't like terms

1 year ago
OpenStudy (iwanttogotostanford):

ok, so what would i do?

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (iwanttogotostanford):

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (freckles):

yep

1 year ago
OpenStudy (iwanttogotostanford):

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?

1 year ago
OpenStudy (freckles):

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

1 year ago
OpenStudy (freckles):

i need to take a short break

1 year ago
OpenStudy (iwanttogotostanford):

ok, thanks!

1 year ago