contestada

Consider the functionf:R→Rdefined viaf(x) =|x|.(a) Give a functiongwith domainRsuch thatg◦fis one-to-one, or describe why it is not possible.(b) Give a function with domain such that◦fis onto, or describe why it is not possible.(c) Give a functiongwith rangeRsuch thatf◦gis one-to-one, or describe why it is not possible.(d) Give a functiongwith rangeRsuch thatf◦gis onto, or describe why it is not possible.

Respuesta :

Answer:

(a) Is not possible

(b) It is possible

(c) It is possible

(d) Is NOT possible.

Step-by-step explanation:

(a)

Is not possible, notice that for any function [tex]g[/tex]  such that

[tex]g : \mathbb{R} \rightarrow \mathbb{R}[/tex]

you would have that

[tex](g\circ f)(x) = g(f(x)) = g(|x|)[/tex]

And for, lets say -3,3 you have that

[tex]g(|-3|) = g(|3|) = g(3)[/tex] therefore is not possible to find a function that is one to one.

(b)

It is possible. Take the following function

[tex]g(x) = x\sin(x)[/tex] since [tex]\sin[/tex] is periodic it will take positive and negative numbers and if you multiply by [tex]x[/tex] each period will become bigger and bigger.

(c)

It is possible. Take the function

[tex]g(x) = \sqrt{x}[/tex]

Then

[tex](f \circ g )(x) = | \sqrt{x} | = \sqrt{x}[/tex]    and [tex]\sqrt{x}[/tex] is one to one.

(d)

It is NOT possible because   [tex](f\circ g)(x) = f(g(x)) = |g(x)|[/tex] and that will always be positive.