# Thread: determining whether this function is one to one or not

1. ## determining whether this function is one to one or not

how do you determine whether this function is one to one or not?
y = x^3 - 4x^2 + 2

f(a) = f(b), a=/=b
a^3 - 4a^2 +2 = b^3 - 4b^2 +2
a^3 - 4a^2 = b^3 - 4b^2 (subtract 2 from both sides)
a^3 - b^3 - 4a^2 + 4b^2 = 0
(a - b)(a^2 + ab + b^2) - 4(a^2 - b^2) = 0
(a - b)(a^2 + ab + b^2) - 4(a + b)(a - b) = 0
(a - b)(a^2 + ab + b^2 - 4a - 4b) = 0

i'm not sure on what to do next. this is where i'm stuck. what i'm supposed to do is solve for the two factors. for example the first factor results in a = b. but i need to somehow solve for the second factor so i can see whether my two solutions contradict with the statement a=/=b or not. i've been using the above method for all of my homework problems but it doesn't seem to work on this particular problem. my teacher only showed us this way of doing it. any suggestions?

2. This is fairly easy if one can use calculus, but I took note that this post is in the precalculus section. I tried playing with the algebra of your attempt, but couldn't make anything break loose ... and then I noticed that f(0) = f(4).

3. Originally Posted by skeeter
This is fairly easy if one can use calculus, but I took note that this post is in the precalculus section. I tried playing with the algebra of your attempt, but couldn't make anything break loose ... and then I noticed that f(0) = f(4).
thanks for trying it haha. i would like to ask how you came up with
f(0) = f(4). are those just random numbers or did you determine them somehow?

4. Originally Posted by oblixps
i would like to ask how you came up with
f(0) = f(4). are those just random numbers or did you determine them somehow?
I do not mean to answer for Skeeter.
However, this is a setup question.
Consider, $\displaystyle f(x)=x^3-ax^2+2\;,\; a\not=0$ then $\displaystyle f(0)=f(a)$.