Define f : R ⇒ R by f(x) = (x^2) - 4x + 6 . What is the range of f ? Is f onto? Explain.?

a) What is the range of f ?


b) Is f onto? Explain.


c) Is f one-to-one? Explain.

Define f : R ⇒ R by f(x) = (x^2) - 4x + 6 . What is the range of f ? Is f onto? Explain.?
(a) The range of f is determined by finding all possible y such that


x^2 - 4x + 6 = y


can be solved for real x.


We have


x^2 - 4x +(6-y) = 0.


This has real solutions if and only if the discrimant ("b^2-4ac") is greater than or equal to zero,


i.e. 16 - 4.1.(6-y) %26gt;= 0,


i.e. 16 - 24 + 4y %26gt;= 0,


=%26gt; y %26gt;= 2.


So the range of f is {x in R : x %26gt;= 2}.





(b) f is not onto using the result in (a), because there are real values which cannot be mapped onto, i.e. all real values less than 2.





(c) Suppose f(a) = f(b), for a,b in R.


Then a^2 - 4a + 6 = b^2 - 4b + 6


=%26gt; a^2 - b^2 = 4(a-b)


=%26gt; (a-b)(a+b) = 4(a-b)


=%26gt; (a-b) (a+b-4) = 0.


This implies that a = b or a=4-b.


The possibility a = 4 - b means that f(a) = f(b) can be satisified with a not equal to b (except for a = 2, when a = b is definitely implied), i.e. every point in the range is mapped onto by two different values (except f(2) = 2).


Hence, f is not one-to-one.
Reply:SInce it is a quadratic with the coefficient of the squared term positive, this equation describes a parabola that opens upward. The lowest point will be the vertex, which is at (2,2). This is found by completing the square





x^2 - 4x + 4 + 6 - 4 = (x-2)^2 + 2





Therefore, the range is all real numbers greater than or equal to 2. Since the range is not all real numbers, the function is NOT onto. It is also not 1-1 as there are two sides to the parabola.