Question

f(x)=2x+3, xE [-1,2]

f^-1(x)= (x-3)/2

What is the domain of f^-1(x)?

Answer 1

\[f(x)=2x+3, x \epsilon [-1,2]\]
\[f^{-1}(x)=\frac{ x-3 }{ 2 }\]
What is the domain of \[f^{-1}(x)\]

Answer 2

can you help me out @zepdrix

Answer 3

Mmm the question is confusing D:

It seems like the domain should be all real numbers, just like f(x).
But why did they give us \(\large x\in[1,2]\), does that have something to do with it? +_+ grr

Answer 4

The domain of f is {-1,2} So the ordered pairs of f(x) are (-1,1) and (2,7) In the inverse function, the ordered pairs are reversed so the domain of the inverse is the range of the original function which is {1,7}

Answer 5

thanks @Mertsj

Answer 6

yw

Answer 7

what about this one @Mertsj

Answer 8

i thought the domain was (-infinity,3)U(3,infinity)

Answer 9

The domain of the f is all reals except 1

Answer 10

Did you find the inverse function?

Answer 11

i think the inverse is (x-2)/(x+3)

Answer 12

Yes. I agree.

Answer 13

so domain is anything but -3

Answer 14

Except the original function has no range value for x = 1. So did you look at the graph to see if there is a vertical asymptote there?

Answer 15

yes, it seems that x cant equal 3 but all other real numbers

Answer 16

thx @Mertsj

Answer 17

yes. I agree.

Answer 18

yw