Check the injectivity and surjectivity of the given functions: f: R → R given by f(x) = x2
f: R → R is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.