@5: f(x)=x^2+2x-3 is not a 1-1-function, polynomials with the highest power being an even one aren't if the domain is the whole real line

the easiest way is to visualise the graph and remember the horizontal line test, see http://www.mathwords.com/o/one_to_one_function.htm
the cut-out posted by 'khemist' is correct but in this case you will have to find a=/=b with f(a)=f(b)
how to find one of those points without sketching the graph or trying plenty numbers?
when is x^2+2x=0? clearly for x=0 but also for x=-2, so you have found two values of x where f(x) is the same: f(0)=f(-2)=-3

@6: the degree of a polynomial is the highest power of variable, ie

3 is the degree of f(x)=-11x^3+3x^2-x+1

and since the leading coefficient is the constant associated with the term of the highest power, it is -11 in this case

how much fg

if someone does a lot of work for you and you want to show your appreciation that's fine,
but you should know that there is no need to pay fg to people helping here voluntarily
as 'Azrad' already indicated in post#10

btw, all questions clear now? and were are the 'many more problems' you claimed you have

got the fg now if u wanna do it

@

you really need to start doing some yourself, else you are not going to learn the stuff

but just as example from your post#16

@7: abs(2*x^2-x-2)=1 should have 4 solutions because the function is quadratic and can have two results, namely 1 or -1, so

2*x^2-x-2=1 -> 2*x^2-x=3 one solution is easy to spot: x=-1 and the other (after a little thinking) is x=1.5

&

2*x^2-x-2=-1 -> 2*x^2-x=1 here the easy solution is x=1 and the other is x=-0.5

ie the four solutions are -1, -0.5, 1, 1.5

this^^

I'll do a "hard" one, #5, from post 16:


This post was edited by Azrad on Nov 14 2013 03:59am