Let A = R − {−2}, B = R − {1} and f : A → B be given by the rule f(x) = x + 1 x + 2 . (a) Prove f is one-to-one. (b) Prove f is onto B. (Comment: don’t forget that if given b ∈ B, you construct a such that f(a) = b, you must also show a ∈ A.)

Explanation:In order to prove that f is one-to-one and surjective, we can prove directly that f is bijective by findng [tex] f^{-1} [/tex] . [tex] f(x) = \frac{x+1}{x+2} [/tex]To calculate the inverse of f, first we swap the variables x and y. Then we  write y in values of x.[tex] x = \frac{y+1}{y+2} [/tex][tex] x/(y+2) = y+1 [/tex][tex] xy +2x = y+1 [/tex]On this step it is important to separate what terms contain y and what terms do not. After that we can separate y as a factor[tex] 2x = y+1 -xy [/tex][tex] 2x-1 = y-xy [/tex][tex] 2x-1 = y(1-x) [/tex][tex] y = \frac{2x-1}{1-x} [/tex]Thus, [tex] f^{-1}(x) = \frac{2x-1}{1-x} [/tex] . Note that [tex] f^{-1} [/tex] is well defined en every element of B, because it can be defined on any element different from 1, where the divider is zero. An element of the form [tex] \frac{2x-1}{1-x} [/tex] , for some x, is an element of A because it is different from -2. Lets suppose that there exist an x such as [tex] \frac{2x-1}{1-x} = -2[/tex] , thus 2x-1 = -2(1-x) = -2+2x, from where we conclude that 0 = -1, that is a contradiction. As a result, [tex] \frac{2x-1}{1-x} [/tex] cant be equal to -2.I hope this works for you!