Answer to Question #307094 in Discrete Mathematics for Snave

Question

Discussion Assignment

Let f(x)=\sqrt(x) with f: \mathbb{R} \to \mathbb{R}. Discuss the properties of f. Is it injective, surjective, bijective, is it a function? Why or why not? Under what conditions change this?

Explain using examples.

Answer 1

1) take x=-1, the function is not defined on it.

2) the function does not take the value -1

It follows that the function is not bijective, surjective, injective.

to change this, you need to change the scope and scope

if f: [0,+$\infin$ )->[0,+$\infin$ ) then function is bijective, surjective, injective.

if f:R->[0,+$\infin$ ) then function is only surjective

if f: [0,+$\infin$ )->R then function is only injective