Question

Consider the function f : R → R, x→ f(x) = x^2−3x+2 (a so-called quadratic polynomial).

i) Is f injective, surjective, or both?
ii) Is f : R → [−0.25, [infinity][, x→ f(x) = x^2 − 3x + 2, injective, surjective, or both?
iii) Determine the domain X ⊆ R and the co-domain Y ⊆ R such that f : X → Y , x→ f(x) = x^2 − 3x + 2, is bijective.

Answer 1

The function f(x) = x^2 - 3x + 2 is not injective or surjective. By restricting the domain and co-domain, we can make it bijective.

Step-by-step explanation:

i) The function f(x) = x^2 - 3x + 2 is not injective because it fails the horizontal line test. This means that there exist different inputs that produce the same output. For example, f(1) = f(2) = 0. Therefore, it is not a one-to-one function.

ii) The function f(x) = x^2 - 3x + 2 is not surjective because the range of the function does not include the interval [-0.25, infinity). For example, there is no x such that f(x) = -0.5. Therefore, it does not map onto the specified range.

iii) To make f(x) = x^2 - 3x + 2 bijective, we can restrict the domain X to [1, infinity) and the co-domain Y to [0, infinity). This way, each element of X has a unique element in Y and vice versa, making it a one-to-one correspondence.