Question

If G(x) = 3x + 1, then G -1(1) is
answer choices


4, 1/3
,0

Answer 1

So first, if you want to find \[G^{-1}(1)=? \]
Then all you real need to do is find for what value ? that G(?)=1.


That is one way to do it.

You could actually start by finding the inverse of G.


So do you know how to solve the following equation for ? .
G(?)=1

Answer 2

Find G(?) first.
Just take the G(x)=3x+1
and replace all the x's with ?'s

Answer 3

Then solve G(?)=1 for ?

Answer 4

This is the question

Answer 5

right.
So you know how to find G(?) ?

Answer 6

no

Answer 7

Do you see the G(x)=3x+1 thing?

To find G(?) replace all the x's in that above function with ?'s

Answer 8

alright
G(?)=3?+1

Answer 9

now solve G(?)=1 for ?

Answer 10

replace G(?) with 3*?+1
like so
G(?)=1
3*?+1=1

Answer 11

its 1/3 right

Answer 12

solve that equation for ?

Answer 13

0

Answer 14

right
\[\text{since } G(0)=1 \text{ then } G^{-1}(1)=0\]

Answer 15

thanks

Answer 16

wait

Answer 17

Given that F(x)=x 2+2, evaluate F(2).


0

2

6

Answer 18

F(2) means to replace the x in F(x)=x 2 +2 with 2.

Answer 19

sorry wrong one

Answer 20

@myininaya

Answer 21

replace the x with what it says to replace it with

Answer 22

then evaluate

Answer 23

its 6

Answer 24

2^+2=6

Answer 25

Just like when I told you to evaluate G(?) given G(x)=3x+1
you did G(?)=3*?+1



so if F(x)=x^2+2 then F(2)=2^2+2
right

Answer 26

well do you see any multiplication or addition?

Answer 27

no

Answer 28

so that means you can eliminate what choices?

Answer 29

both of those

Answer 30

what choices are left?

Answer 31

distributive and sym

Answer 32

do you know the distributive property?

Answer 33

can you do me a favor

Answer 34

What is the favor?

Answer 35

can you sign into this site under my name and do this for me ill pay or you could be kind hearted and finish this moduel for me your a math wiz

Answer 36

That is cheating though. I cannot do your work. It is not helping you. It would only be damaging you for future tests, quizzes, and other homework.
Please look at the CoC http://openstudy.com/code-of-conduct and the terms and conditions http://openstudy.com/terms-and-conditions .
These are both things you agreed to when you signed up on OpenStudy.

Answer 37

youre right i need to do this but damn seems like nobody wants to help anybody on here

Answer 38

I was trying to help you...

Answer 39

yeah but youre gonna leave sonner or later

Answer 40

wrong one

Answer 41

Here is a hint:
\[x^0 =1 \text{ note: assuming } x \neq 0\]

Answer 42

its 1

Answer 43

yep
anything with the power 0 is 1
except for when you have 0^0

Answer 44

you are suppose to go under the assumption a+b is not 0
so (a+b)^0 would be 1

Answer 45

hint:
\[\frac{1}{x^n}=x^{-n}\]

Answer 46

\[\frac{1}{(something)^n}=(something)^{-n}\]

Answer 47

1/6?

Answer 48

I'm seeing this problem:
\[\frac{1}{(xy)^1}\]

Answer 49

Are you doing a different one...

Answer 50

yes thats the right problem

Answer 51

look at what I wrote above what is the something and what is the n value?

Answer 52

compare 1/(xy)^1
to 1/(something)^n

Answer 53

determine what something is and what the n is and then replace it on the side of that equation i wrote

Answer 54

\[\frac{1}{(something)^n}=(something)^{-n}\]

Answer 55

\[\frac{1}{(xy)^1}=( ? )^{- *}\]

your job
is to replace that ? and to replace the *

Answer 56

1x and i dont know the last part is it -2

Answer 57

I see no 2's in 1/(xy)^1

Answer 58

what happen to the y

Answer 59

yeah youre right

Answer 60

wym

Answer 61

\[\frac{1}{(xy)^n}=\frac{1}{(something)^n}\]
Compare both sides of this equation
What would the something have to be and what would the n have to be for both of the sides to be the same as each other.
what is in the something's place
what is in the n's place

Answer 62

1