Answer to Question #305791 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

$f'(x)=\frac{1}{2\sqrt{x}}>0, \ for\ x\in[0,+\infin)$

f is not surjective and injective since the range and domain are [0,+$\infin$ )

for f to be a bijection is necessary and sufficient, f must be injection and surjection=>

f is not bijective

if f:[0,+$\infin$ )->[0,+$\infin$ )

f-increasing function сonsequently exist 

$f^{-1}$

f- is bijective=>f is injective and surjective