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!

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

answer

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step-by-step explanation: