Thread: discrete math function

determine whether the function f(x)^2-3(x)+1 is one-to-one, onto, or both. Prove your answer. The domain and codomain if f(x) is the set of all real numbers.

f(x) = 3x^2 -3x + 1

the function is onto Y, but not one-to-one:

because 1 is a real number and

f(1) = 3(1)^2-3(1)+1 = 1

and 0 is a real number

f(0) = 3(0)^2-3(0)+1 = 1

there exists two different elements in x that map to a single element in y; therefore, f is not one-to-one.

the function is onto Y, since all real numbers of Y coexist with all real numbers of X.



OK, my question is did I correctly prove that

1. the function is one-to-one?

2. how do I prove the last one is true algebraically?

I assume its true since the domain is (-00, 00) and range is(-00, 00)?

thanks!

Originally Posted by robasc

determine whether the function f(x)^2-3(x)+1 is one-to-one, onto, or both. Prove your answer. The domain and codomain if f(x) is the set of all real numbers.

f(x) = 3x^2 -3x + 1

the function is onto Y, but not one-to-one:

because 1 is a real number and

f(1) = 3(1)^2-3(1)+1 = 1

and 0 is a real number

f(0) = 3(0)^2-3(0)+1 = 1

there exists two different elements in x that map to a single element in y; therefore, f is not one-to-one.

the function is onto Y, since all real numbers of Y coexist with all real numbers of X.


"Coexist"? That is not a mahtematical term. The function is not onto the real numbers, not even close: that quadratic's

discriminant is , and this means the function is always positive or always negative...

Tonio



OK, my question is did I correctly prove that

1. the function is one-to-one?

2. how do I prove the last one is true algebraically?

I assume its true since the domain is (-00, 00) and range is(-00, 00)?

thanks!

.

I see now since this is a parabola it does not even map to any negative values in y, but i am not following the discriminant ?

I do not think I am familiar with discriminant. I am a serious newbie at this stuff.

please explain how to do this?