Question

If
f(x)=3x+1 and f-1 (inverse)=x-1 over 2
Then f(2)=
1/3
1
7
please help with functions and relations:inverses!!

Answer 1

\[\Large f(\color{royalblue}{x})=3(\color{royalblue}{x})+1\]
\[\Large f(\color{royalblue}{2})=3(\color{royalblue}{2})+1\]

Answer 2

Function notation can take a little while to get comfortable with :)
Do the colors help at all?

Answer 3

For f(2) we're letting x=2 in all the places we would normally have an x.

Answer 4

@zepdrix
yes it really does...what made me confuse is I thought I was suppose to plug it into the x-1 over two

Answer 5

I'm not sure why they gave us that piece of information (about the inverse function).
Are there other questions to follow that use these two functions maybe?
Others asking about \(\large f^{-1}(2)\) or something maybe?

Answer 6

yes I don't either, it really threw me off....so it was the same question how would I find f-1(7) same idea

Answer 7

\[\Large f^{-1}(\color{teal}{x})\quad=\quad \frac{(\color{teal}{x})-1}{2}\]
\[\Large f^{-1}(\color{teal}{-7})\quad=\quad \frac{(\color{teal}{-7})-1}{2}\]

Answer 8

If this is supposed to be the inverse of the first f they provided, the denominator should be a 3, not a 2. Hmmm.

Answer 9

Yes it there is a 2, I got it wrong because I had -2

Answer 10

So (-7)-1 over two...I'm not understanding how we are suppose to get two...also another option I have is 22, 2, -2

Answer 11

For the inverse question?\[\Large \frac{-7-1}{2}\quad=\quad \frac{-8}{2}\quad=\quad -4\]That's not one of the options?

Answer 12

Oh lolol :)
You asked for f^{1}(7)... not -7, sorry i misread that. :p

Answer 13

\[\Large f^{-1}(\color{teal}{x})\quad=\quad \frac{(\color{teal}{x})-1}{2}\]
\[\Large f^{-1}(\color{teal}{7})\quad=\quad \frac{(\color{teal}{7})-1}{2}\]

Answer 14

So it looks like we're getting uhhh, 3 yes?\[\Large \frac{7-1}{2}\quad=\quad\frac{6}{2}\quad=\quad 3\]

Answer 15

No I'm sorry there is not a three :/

Answer 16

Can you take a picture of the problem maybe? D:

Answer 17

If
f(x)=3x+1 and f-1= x-1 over 3

then f -1(7) =
22, 2,-2 ???

Answer 18

`Yes it there is a 2,`....
I thought u said the denominator was a 2...

Answer 19

Oh...I'm sorry about that, it was my fault

Answer 20

\[\Large f^{-1}(\color{teal}{x})\quad=\quad \frac{(\color{teal}{x})-1}{3}\]
\[\Large f^{-1}(\color{teal}{7})\quad=\quad \frac{(\color{teal}{7})-1}{3}\]Ok if we have a 3, it should give us one of the answers on the list :D

Answer 21

But yes, that makes a lot of sense the way you set it up...I really got that:)

Answer 22

cool c:

Answer 23

Yep thanks a lot, think I'm ready to take my quiz now:)

Answer 24

Let R={0, 1, 2, 3} be the range of h(x) = x - 7. The domain of h inverse is

{-7, -6, -5, -4}
{7, 8, 9, 10}
{0, 1, 2, 3}
can you please explain this last one, I got the last answer and it was right but I just took a wild guess on it:/

Answer 25

When we take the inverse of a function, the `domain` and `range` are exchanged.

So if the `domain` of \(\large h(x) \)is R, then the `range` of \(\large h^{-1}(x)\) will also be R.

Answer 26

Here's a way to think of it maybe, ummm....


If f(x) has `domain` D and `range` R,

then f inverse will have `domain` R and `range` D.

Answer 27

@zepdrix
I understand that, but like how come the answer stay the same...Oh so the range would be (7, 8,9,10) then when you interchange them...it would be like 7-7=0, 8-7=1,9-7=2, and 10-7=3 so that the answer would just stay the same??? Did I understand that right?

Answer 28

So for h(x):
R={0, 1, 2, 3}

If we plug these in for h(x) we get a domain of:
D={-7, -6, -5, -4}
-----------------------------------------
So for h(x):
Domain={-7, -6, -5, -4}
Range={0, 1, 2, 3}
-----------------------------------------
For h inverse, we switch them:
Domain={0, 1, 2, 3}
Range={-7, -6, -5, -4}

Answer 29

h(x) Range={0, 1, 2, 3}

h inverse Domain={0, 1, 2, 3}

right? :o

Answer 30

@zepdrix
Yes that's right:D and I get it now but I'll study more before I take my quiz...anyways thank you so much for the clarifications it helped a lot

Answer 31

\c:/ np