Question

Are my answers correct? Attached files. PLEASE CHECK I'M TURNING IT IN.

Answer 1

Ok hold on a sec @ohlookidontunderstandanotherthing

Answer 2

okay c:

Answer 3

@ohlookidontunderstandanotherthing

Answer 4

Sorry I was doing something! Give me a sec to look into this question :D

Answer 5

Thanks @ohlookidontunderstandanotherthing

Answer 6

a) is correct

on b) I can't make out who is f(x) and who is g(x) though

are both f(x) and g(x) (1/2x)+1?

Answer 7

No, I changed the f(x) to g(x) in b

Answer 8

because that is what it askes me to do. How about c?

Answer 9

^^ jdoe0001 is far more intelligent then I am. They'll answer better xD

Answer 10

hmmmm I see g(x) is the one obtained from a)

Answer 11

no, it's the inverse from a.

Answer 12

right

Answer 13

so I'm correct?

Answer 14

so.... did you show that.... I see that part there... I see the writing....

Answer 15

What do you mean?

Answer 16

Well, yes I did show it. It's above the answer

Answer 17

hmmm I see a)'s answer.... I don't see b)'s though

Answer 18

there is like a crossout with the pencil, it's dark, and beside it is the explination.

Answer 19

well... I see that... not very readable anyhow
lemme do say f( g(x) )

Answer 20

\(\bf f(x)=y=\cfrac{1}{2}x+1\qquad g(x)=2(x-1)
\\ \quad \\
f(\ g(x)\ )=\cfrac{1}{\cancel{ 2 }}[\cancel{ 2 }(x\cancel{ -1 })]\cancel{ +1 }\to x\)

Answer 21

I think that's what I have

Answer 22

Could you tell me if c is correct?

Answer 23

k

Answer 24

for one thing the graph looks like http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoMS8yKXgrMSIsImNvbG9yIjoiI0UwMzQzNCJ9LHsidHlwZSI6MCwiZXEiOiIyKHgtMSkiLCJjb2xvciI6IiMxRjFGQ0YifSx7InR5cGUiOjEwMDAsIndpbmRvdyI6WyItNC4yMTk5OTk5OTk5OTk5OTkiLCI4Ljc4MDAwMDAwMDAwMDAwMSIsIi0xLjg2MDAwMDAwMDAwMDAwMDMiLCI2LjE0Il19XQ--

Answer 25

so notice the green line.... that is the y = x line

http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIoMS8yKXgrMSIsImNvbG9yIjoiI0UwMzQzNCJ9LHsidHlwZSI6MCwiZXEiOiIyKHgtMSkiLCJjb2xvciI6IiMxRjFGQ0YifSx7InR5cGUiOjAsImVxIjoieCIsImNvbG9yIjoiIzRCRjIzNSJ9LHsidHlwZSI6MTAwMCwid2luZG93IjpbIi00LjIxOTk5OTk5OTk5OTk5OSIsIjguNzgwMDAwMDAwMDAwMDAxIiwiLTEuODYwMDAwMDAwMDAwMDAwMyIsIjYuMTQiXX1d

the "blue" is reflecting the "red" over the green line, or y = x line

Answer 26

and inverse functions graph, do reflect each other over the y=x line

Answer 27

Thank You so much *-*