Question 1131828
{{{f(x) = x^2 + 1}}} is {{{not}}} injective (one-to-one) onto {{{R}}}
Injective means we won't have two or more "{{{x}}}"s pointing to the same "{{{y}}}".
In other words there are two values of {{{x}}} that point to one {{{y}}}.

 Function is said to be injective or one-to-one if every element in the range is an image of at most one element from the domain.
An injective function is called a one-to-one function.
Since {{{x}}} is on the right side of the equation, switch the sides so it is on the left side of the equation.
{{{x^2+1=y}}}
Add {{{-1}}} to both sides of the equation: {{{x^2=y-1}}}
Take the square root of both sides of the equation to eliminate the exponent on the left side.
{{{x}}}=±{{{sqrt(y-1)}}}
The complete solution is the result of both the positive and negative portions of the solution.

{{{x=sqrt(y-1)}}},{{{x=-sqrt(y-1)}}}
There is more than y value for some x values, which means that y=x^2+1 is not an equation of a function.


{{{f(x) = x^2 + 1}}} is not surjective onto {{{R}}}
A function {{{f(x)}}}  (from set A to B) is surjective if and only if for every {{{y}}} in B, there is at least one {{{x}}} in A such that {{{f(x) = y}}}, in other words  {{{f(x)}}}  is surjective if and only if {{{ f(A) = B}}}.
In simple terms: every {{{B}}} has some {{{A}}}.

Surjective (Also Called "Onto")means that every "{{{y}}}" has at least one matching "{{{x}}}" (maybe more than one).
There won't be a "{{{x}}}" left out.  

But, 
domain is {{{R }}} (all real numbers) and 
range (codomain) is 
{ {{{f(x)}}} element {{{R}}} : {{{f(x) >= 1}}} }

So, in this function all the negative values in the codomain of {{{f(x)}}} do not have any corresponding {{{x}}} values in the domain of {{{f(x)}}}