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 .

To calculate the inverse of f, first we swap the variables x and y. Then we  write y in values of x.

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

Thus, . Note that 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 , for some x, is an element of A because it is different from -2. Lets suppose that there exist an x such as , thus 2x-1 = -2(1-x) = -2+2x, from where we conclude that 0 = -1, that is a contradiction. As a result, cant be equal to -2.

I hope this works for you!

10x

step-by-step explanation:

the greatest common factor 30 and 20 is 10

5 is a common factor of 30 and 20 but 10 is also a factor and 10 is greater than 5 so 5 will definitely not be apart of our answer

now the most amount of x's you factor our is just 1 factor of 1

so 10x the gcf of the terms in the polynomial

£56

step-by-step explanation: