Classify `f: R->R` , defined by `f(x)=1+x^2` as injection, surjection or bijection.

Explanation

Injection condition:

Let `x` and `y` be any two elements in domain (`R`), such that `f(x) = f(y)`.

`f(x) = f(y)=> 1 + x^2 = 1 + y^2=> x^2 = y^2=> x = ± y` Therefore, f is not an injection.

Surjection test:

Let `y` be any element in the co-domain (`R`), such that `f(x) = y` for some element `x` in `R` (domain).

`f(x) = y`

`1 + x^2 = y`

`x^2 = y − 1`
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