• Mar 30th 2009, 04:20 PM
fumanchu
Hi - I guess this is easy but I am having trouble solving it. I am supposed to find the inverse of the quadratic function

y = x^2 - 2x +1

Thanks in advance for any help, I am not sure if I can just factor it? The answer is supposed to be y = 1 +- srqrt X
• Mar 30th 2009, 04:28 PM
EmpSci
Quote:

Originally Posted by fumanchu
Hi - I guess this is easy but I am having trouble solving it. I am supposed to find the inverse of the quadratic function

y = x^2 - 2x +1

Thanks in advance for any help, I am not sure if I can just factor it? The answer is supposed to be y = 1 +- srqrt X

You start by substituting x with y:

x = y^2 - 2y + 1. Now we want to express y in terms of x. So, factor the polynomial:

x = (y-1)^2.

Therefore, y-1 = +-SQRT(x), so y = 1 +- SQRT(x).
• Mar 30th 2009, 04:31 PM
fumanchu
Thanks sorry I didn't get it at first.

Do you know of a place that explains why the x ends up as + or - sqrt(x) when you take the square root. I understand that is the answer but I don't really get why. Sorry again.
• Mar 30th 2009, 08:56 PM
Gaurav
Quote:

Originally Posted by fumanchu
Do you know of a place that explains why the x ends up as + or - sqrt(x) when you take the square root. I understand that is the answer but I don't really get why. Sorry again.

this is so because whwn u mutiply a neg. no. by itself(-a*-a) the result will come positive.
and if u multiply a positive no. with itself(a*a) the result will be positive.
so it cant be made out if the sqrt was originally posit. or neg.. so to be on the safer side we take both +&-.
• Mar 31st 2009, 08:27 AM
stapel
Quote:

Originally Posted by fumanchu
Do you know of a place that explains why the x ends up as + or - sqrt(x) when you take the square root.

You get the "plus-minus" from the Quadratic Formula or, if you prefer, from completing the square. (But that's the hard way!)

(Wink)
• Mar 31st 2009, 02:58 PM
Referos
Wait, that can't be the inverse function. When you have the plus-minus sign, you have a one-to-many relation, so it is not a function! The exercise must give the original's function domain and codomain if you are expected to find its inverse.
• Apr 1st 2009, 06:11 AM
stapel
But the exercise, as originally posted, did not ask for an inverse function; it asked only for "the inverse". The "plus-minus" relation fulfills that requirement. :D
• Apr 2nd 2009, 08:22 AM
Referos
Well, I suppose this is just semantics and is really irrelevant, but the OP did say "find the inverse of the quadratic function" and the inverse of a function must also be a function. So it must be asking the inverse function.
• Apr 3rd 2009, 04:00 PM
stapel
Quote:

Originally Posted by Referos
...the inverse of a function must also be a function.

No; only some functions are "invertible"; only some functions have inverses which are also themselves functions. Usually, the inverse is just a "relation". (Wink)