1. ## Bijection

Determine if this is a bijection and find the inverse function.f : R − {−2} → R − {1} where f(x) = (x + 1)=(x + 2).

I have the ans sheet: It's a bijection with inverse f-1(x) = (2x 1)=(1 x).

2. ## Re: Bijection

Originally Posted by ZGOON
Determine if this is a bijection and find the inverse function.f : R − {−2} → R − {1} where f(x) = (x + 1)=(x + 2).
I have the ans sheet: It's a bijection with inverse f-1(x) = (2x − 1)=(1 − x).

Do you mean $f(x)=\frac{x+1}{x+2}~?$

3. ## Re: Bijection

Sorry for the error.

f : R − {−2} → R − {1} where f(x) = (x + 1)/(x + 2).

inverse f-1(x) = (2x − 1)/(1 − x).

4. ## Re: Bijection

Originally Posted by ZGOON
Sorry for the error.
f : R − {−2} → R − {1} where f(x) = (x + 1)/(x + 2). inverse f-1(x) = (2x − 1)/(1 − x).

If $t\in\mathbb{R}\setminus\{1\}$ is it possible to solve $t=\frac{x+1}{x+2}~?$ (i.e. get $x$ in terms of $t$)

5. ## Re: Bijection

How you get x-2???

It is function since every x in the domain has an image. 1-1: f(a) = f(b) implies
(a+1)/(a+2) = (b+1)/(b+2). This yields a = b. Onto: Let y ∈ R−{1}. Set f(x) = y.
Solving, we get x = (2y − 1)/(1 − y). Thus If (2y − 1)/(1 − y) = −2, then we have 1 = 2
which is impossible. Thus (2y − 1)/(1 − y) is never equal to −2 and is thus a preimage.
The above calculations also show that f-1(x) = (2x − 1)/(1 − x).

6. ## Re: Bijection

Originally Posted by ZGOON
How you get x-2???
Typo