Show that the function `f: R ->R` is given by `f(x)=1+x^2` is not invertible.

Explanation

Given, function `f: R arrow R` such that `f(x)=1+x^{2}`,


Let `A` and `B` be two sets of real numbers.


Let `{x}_{1}, {x}_{2} in {A}` such that `{f}({x}_{1})={f}({x}_{2})`.



`Rightarrow 1+{x}_{1}^{2}=1+{x}_{2}^{2} Rightarrow {x}_{1}^{2}-{x}_{2}^{2}=0 Rightarrow({x}_{1}-{x}_{2})({x}_{1}+{x}_{2})=0`



`Rightarrow {x}_{1}=pm {x}_{2}`. Thus `{f}({x}_{1})={f}({x}_{2})` does not imply that `{x}_{1}={x}_{2}`.


For instance, `f(1)=f(-1)=2`, i.e. , two elements `(1,-1)` of A have the same image in B. So, `f` is many-one function.
...