Respuesta :
Answer:
Check below
Step-by-step explanation:
1. Definition for intervals
[tex](a,b)=\left \{ x\in\Re : a<x \:and \:x<b \right \}\\(a,b]=\left \{ x\in\Re : a<x \:and \:x\leq b \right \}[/tex]
2. Functions
1) [tex]\Re \rightarrow \Re \\ f(x)=x[/tex]
Let's perform graph tests.
That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.
2)[tex]\Re\geq0\rightarrow\Re , f(x)=x+1\\[/tex]
Injective, surjective and bijective.
Injective: a horizontal line crosses the graph in one point.
3)[tex]f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)[/tex]
The cosine function is not injective, bijective nor surjective.
4)[tex]f:\Re\rightarrow\Re \:f(x)=ex[/tex]
Since e, is euler number it's a constant. It's also injective, surjective and bijective.
5) Quite unclear format
[tex]6) \:f:\Re\rightarrow(0,\infty), f(x) =ex[/tex]
Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.
[tex]7) f:\Re\geq 0 \rightarrow \Re\geq0, f(x) =x^{4}[/tex]
Not injective nor surjective therefore not bijective too.
[tex]8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}[/tex]
[tex]f(-1)=1, f(2)=4, f(-3)=9[/tex]
Injective (one to one), Surjective, and Bijective.
[tex]10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.[/tex]
Surjective.
[tex]11.f:R\geq 0[-1,1], f(x) = 0\\[/tex]
Surjective
12.f: US Citizens→Z, f(x) = the SSN of x.
General function
13.f: US Zip Codes→US States, f(x) = The state that x belongs to.
Surjective
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