# Polynomial Functions.

1. Oct 31, 2011

### mtayab1994

1. The problem statement, all variables and given/known data

f:]1,+∞[→]2,+∞[
x→ 2x/(x-1)

2. Relevant equations
Prove that f is injective and serjective.

3. The attempt at a solution

I already proved that it's injective by stating the injectivity law:
for every (a,b)ε]1,+∞[: f(a)=f(b) implies a=b

so: 2a/(a-1)=2b/(b-1) entails: 2ab-2b=2ab-2a entails -2b=-2a entails a=b

Can anyone please tell me how to prove that its serjective?

2. Oct 31, 2011

### LCKurtz

Sure. If y is in ]2,+∞[, calculate what x in ]1,+∞[ gives f(x) = y.

3. Oct 31, 2011

### Ray Vickson

In your own words, what does it mean to say that f is surjective? (That is sUrjective, not sErjective!) Turn that verbal statement into an equation and then work on the equation, to see what conclusions you can make, or else use some known, general properties to get a conclusion.

RGV

4. Oct 31, 2011

### mtayab1994

i got it
f(x)=y
y=2x/x-1 equivalence y(x-1)=2x equivalence yx-2x-y=0
now we find Δ
Δ=4+4y^2
since Δ≥0 therefore there is some solution to this equation and therefore f is serjective.

5. Oct 31, 2011

### Ray Vickson

What does $\Delta = 4 + 4y^2$ have to do with anything here? Anyway, you are still spelling surjective incorrectly.

RGV

6. Oct 31, 2011

### mtayab1994

Well my teacher stated that if we find that Δ≥0 then therefore f is surjective and btw my first language is english, but i'm learning overseas in Morocco and all the lessons here are in Arabic, so that's probably the reason why i spelled it wrong.

7. Oct 31, 2011

### Deveno

i believe he's taking the discriminant of a quadratic.

8. Oct 31, 2011

### Ray Vickson

Of course he is; but where is the quadratic equation in this question?

RGV