To be onto, for any integer y, there exists an integer xsuch that y = f(x).Give a counterexample, let y = 20, which is aninteger. If f(x) = 20, then x =20which isnot an integer. To be onto, for any real number y, there exists a realnumber x such that y = f(x).Let y be a real number and let x =Sinceis a real number,mustbe a real number.Henceis a real numberIt follows that f(x) = 2x + 7 =Since f(x) =where x is a real number, f isonto.

To be onto for any real number y, there exists a real number x such that y = f(x). Let y be a real…

Answered: To be onto, for any integer y, there…

To be onto, for any integer y, there exists an integer x such that y = f(x). Give a counterexample, let y = 20, which is an integer. If f(x) = 20, then x = 20 which is not an integer.

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

To be onto, for any integer y, there exists an integer x
such that y = f(x).
Give a counterexample, let y = 20, which is an
integer. If f(x) = 20, then x =
20
which is
not an integer.

To be onto, for any real number y, there exists a real
number x such that y = f(x).
Let y be a real number and let x =
Since
is a real number,
must
be a real number.
Hence
is a real number
It follows that f(x) = 2x + 7 =
Since f(x) =
where x is a real number, f is
onto.

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