Kindly explain the beloe problem how it is Bijective function

1 The fnction f: R -R ,defined by f(x) =2x+1

2 f:N to N defined by f (n) =2n is one-one but not onto

3 Function R to R defined by f ( x) =x square

Here what is the domain set,image set and codomain Kindly explain taking some elements

(1)
f(x) = 2x+1
let $x_1 , x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$
$$\begin{gathered} 2x_1+1=2x_2+1 \\ \Rightarrow x_1=x_2 \end{gathered}$$ 
therefore the function is one-one.
now $y=2x+1\Rightarrow x=\frac{y-1}{2}$ which is defined for all $y\in R$
thus the function is onto.
therefore f(x) =2x+1  is bi-jective function.
(2)
f(n)=2n
​let $n_1 ,n_2\in N$ such that $f\left(n_1\right)=f\left(n_2\right)$
$2n_1=2n_2\Rightarrow n_1=n_2$ therefore the function is one-one.
since for all the odd numbers in N , we do not have the pre-image in N.
only the pre-image of even number exist , therefore the function is not onto.
(3)
the given function is $f:R\to R; f\left(x\right)=x^2$.
since it is defined for all real number ,therefore domain = R.
co domain is R.
and the range is positive real number $R+$

hope this helps you