Thread: Verify that f is a function? Something went wrong in my equation solving or?

Greetings guys
I'm solving an assignment for class tomorrow and have to verify that this relation is a function.
The relation f in real is given by
xfy <=> (y(2x-3)-3x=y(x^2-2x)-5x^2)

AFAIK:
Function: A function is a set of ordered pairs in which each x-element has only ONE y-element associated with it.

So:
I decided to isolate y in the relation:
(y(2x-3)-3x=y(x^2-2x)-5x^2) <=>
2xy-3y-3x=y(x^2-2x)-5x^2) <=>
Divides with y on both sides:
4x-3-3x/y=x^2-(5x^2)/y <=>4x-3-x^2=3x/y - (5x^2)/y
y(4x-3-x^2)=3x-5x^2<=>y=[3x-5x^2]/[4x-3-x^2]

When i do a graph for y. I see that the relation isnt a function, since the relation shares coordinates on x about =[0;2] (*the exact value doesnt matter)

Originally Posted by BruceAlge

When i do a graph for y. I see that the relation isnt a function, since the relation shares coordinates on x about =[0;2] (*the exact value doesnt matter)

What do you mean by "the relation shares coordinates on x"?

I mean that the function fails the vertical line test, which checks wether a relation is a function or not . Meaning that it got several y points for the same x coord.

I am not convinced that it has several points above the same x coordinate. Why do you think so? Maybe it's a side effect of graph plotting in the presence of vertical asymptotes?

I can see that from
]-inf;1[ its sligtly positive (besides from the small part on ]0;0.5[
[1;3] it goes negative
]3;infi+[ it goes positive.
Is that enough to desribe the relation as a function? I am still not quite convienced there isnt dual spots around 1 and 3...

Edit: If i solve the equation for 1 & 3. I get division by 0. Therefor its a function? As there is only uniques x & y's?

You showed that xfy <=> y(4x - 3 - x^2) = 3x - 5x^2. I think in the proof you first divided and then multiplied both sides by y, which risks breaking the equivalence (multiplying both sides by the same expression can provide new roots to an equation). However, the equivalence above is easy to show without dividing or multiplying both sides.

So, for each x such that x^2 - 4x + 3 is not zero, y has to be (5x^2 - 3x) / (x^2 - 4x + 3). For such x there exists a unique y such that xfy. If x^2 - 4x + 3 = 0, i.e., if x = 1 or x = 3, then 5x^2 - 3x is not zero. Therefore, for no y it is the case that xfy. Thus, f is a function, though not total (not defined at x = 1, 3).

If the right-hand side were 0 for x = 1 or x = 3, then xfy would hold for any y. In that case, f would not be a function.

I think you are right Emakarov. Thanks alot man. Conclusion: We got a function. Its domain is Dom(f) = R\{1,3}. However I'm unsure about the part wether its surjective / injective Lets look on the definitions: A surjective function is a function whose image is equal to its codomain. Equivalently, a function f with domain X and codomain Y is surjective if for every y in Y there exists at least one x in X with . Surjections are sometimes denoted by a two-headed rightwards arrow, as in f : X ↠ Y. Symbolically, Let , then is said to be surjective if Does it affect the entire conclusion that {1,3} isnt a part of Dom(f)? et f be a function whose domain is a set A. The function f is injective if for all a and b in A, if f(a) = f(b), then a = b; that is, f(a) = f(b) implies a = b. Equivalently, if a ≠ b, then f(a) ≠ f(b). Symbolically, which is logically equivalent to the contrapositive, Conclusion injective: Its not injective as it got the same y around x=]0;1[ and x = [4;inf+[ Could be veried by adding a horizontal line and plotting its varies interactings with the function?

To check whether f is injective, check x = 0 and x = 3/5. To check whether f is surjective, see if f(x) = -2 for some x.

Thanks man. I figured that about f(0) = f(3/5)= 0 aswell.
Lets check if f(x)=-2.
(3*x-5*x^2)/(4*x-3-x^2)=-2
-8x+6+2x^2=3x-5x^2
7x^2-11x+6=0
d= (-11)^2-4*7*6=-47 => no solutions for f(x)=-2

Could u elaborate on why this matters?

Edit: Nvm think i figured it..
The function f is surjective if it hits everything in the target set, R, but our function doesnt "hit" 2.